PHP Form array has empty values when inserting into MySQL table












1














I'm having an issue inserting php form array values into a MySQL table. Right now the form itself pulls worker information from another table to populate the fields and their values. Right now there are only three workers that would populate those fields, however as more workers get added, there could be as many as 40. When I add just the first worker from the form, all the information inserts normally. However, when I add more than one, the title and employeeId fields are blank and I can't figure out why. Any help would be greatly appreciated.



Here is the form:



<form method = 'POST' action = 'addworkers.php'>
<?php
$sql2 = "select * from workers where companyId = 1";
$result2 = mysqli_query($conn,$sql2);
$numRows = mysqli_num_rows($result2);
$check = 0;
while($row2 = $result2->fetch_assoc()) {
$employeeId = $row2["id"];
$name = $row2["name"];
echo '<input type="checkbox" name="employeeId" value="' . $employeeId . '">';
echo "$namen";
echo '<input type = "hidden" value="' . $companyId . '" name = "companyId"/>';
echo '<input type = "hidden" value="' . $jobNumber . '" name = "jobNumber"/>';
echo 'Site Title : <input type = "text" name = "title"/><br/>';
}
?>
<input type="hidden" name="count" value="<?php echo "$numRows"; ?>"/>
<input type = 'submit' value = 'SEND'/>
</form>


Then the addworkers.php code



require_once("dbConfig.php");
session_start();
$timestamp = date("Y-m-d");

if (isset($_SESSION['loginname'])) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

foreach($title as $key=>$value){
if (!empty($value)) {
$query = "insert into `Jobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId[$key]', '$jobNumber[$key]','$employeeId[$key]','$value','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}


I changed the echo "Error" line but the output was clear.



It must be a problem with how the array is counted but I'm not sure how to fix it. in the form, if I check the boxes next to each row, all the information is entered into the table properly. If I only check the second and/or third line, it doesn't include the Site Title and the employeeId is reversed.



Here is the output of the form:



<form method = 'POST' action = 'insertworkers.php'>
<input type="checkbox" name="employeeId" value="1">Mike
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="2">Steve
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="3">Roger
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="hidden" name="count" value="3"/>
<input type = 'submit' value = 'SEND'/>
</form>


I also changed the foreach loop to a for loop since the array depth should be the same as all fields will be mandatory



require_once("dbConfig.php");
session_start();
$counter = "".$_POST["count"]."";
$timestamp = date("Y-m-d");

if ( isset( $_SESSION['loginname'] ) ) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

for($i=0, $count = count($employeeId);$i<$count;$i++){
if (!empty($employeeId)) {
$query = "insert into `customerJobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId', '$jobNumber','$employeeId[$i]','$title[$i]','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}









share|improve this question
























  • Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
    – Funk Forty Niner
    Nov 21 at 3:03










  • also use error reporting php.net/manual/en/function.error-reporting.php
    – Funk Forty Niner
    Nov 21 at 3:06










  • It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
    – miken32
    Nov 21 at 3:07










  • open to SQL injection attack, dont use this in actual production
    – user10226920
    Nov 21 at 3:10










  • You're only doing a foreach for the one array here, what about the others?
    – Funk Forty Niner
    Nov 21 at 3:11
















1














I'm having an issue inserting php form array values into a MySQL table. Right now the form itself pulls worker information from another table to populate the fields and their values. Right now there are only three workers that would populate those fields, however as more workers get added, there could be as many as 40. When I add just the first worker from the form, all the information inserts normally. However, when I add more than one, the title and employeeId fields are blank and I can't figure out why. Any help would be greatly appreciated.



Here is the form:



<form method = 'POST' action = 'addworkers.php'>
<?php
$sql2 = "select * from workers where companyId = 1";
$result2 = mysqli_query($conn,$sql2);
$numRows = mysqli_num_rows($result2);
$check = 0;
while($row2 = $result2->fetch_assoc()) {
$employeeId = $row2["id"];
$name = $row2["name"];
echo '<input type="checkbox" name="employeeId" value="' . $employeeId . '">';
echo "$namen";
echo '<input type = "hidden" value="' . $companyId . '" name = "companyId"/>';
echo '<input type = "hidden" value="' . $jobNumber . '" name = "jobNumber"/>';
echo 'Site Title : <input type = "text" name = "title"/><br/>';
}
?>
<input type="hidden" name="count" value="<?php echo "$numRows"; ?>"/>
<input type = 'submit' value = 'SEND'/>
</form>


Then the addworkers.php code



require_once("dbConfig.php");
session_start();
$timestamp = date("Y-m-d");

if (isset($_SESSION['loginname'])) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

foreach($title as $key=>$value){
if (!empty($value)) {
$query = "insert into `Jobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId[$key]', '$jobNumber[$key]','$employeeId[$key]','$value','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}


I changed the echo "Error" line but the output was clear.



It must be a problem with how the array is counted but I'm not sure how to fix it. in the form, if I check the boxes next to each row, all the information is entered into the table properly. If I only check the second and/or third line, it doesn't include the Site Title and the employeeId is reversed.



Here is the output of the form:



<form method = 'POST' action = 'insertworkers.php'>
<input type="checkbox" name="employeeId" value="1">Mike
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="2">Steve
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="3">Roger
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="hidden" name="count" value="3"/>
<input type = 'submit' value = 'SEND'/>
</form>


I also changed the foreach loop to a for loop since the array depth should be the same as all fields will be mandatory



require_once("dbConfig.php");
session_start();
$counter = "".$_POST["count"]."";
$timestamp = date("Y-m-d");

if ( isset( $_SESSION['loginname'] ) ) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

for($i=0, $count = count($employeeId);$i<$count;$i++){
if (!empty($employeeId)) {
$query = "insert into `customerJobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId', '$jobNumber','$employeeId[$i]','$title[$i]','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}









share|improve this question
























  • Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
    – Funk Forty Niner
    Nov 21 at 3:03










  • also use error reporting php.net/manual/en/function.error-reporting.php
    – Funk Forty Niner
    Nov 21 at 3:06










  • It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
    – miken32
    Nov 21 at 3:07










  • open to SQL injection attack, dont use this in actual production
    – user10226920
    Nov 21 at 3:10










  • You're only doing a foreach for the one array here, what about the others?
    – Funk Forty Niner
    Nov 21 at 3:11














1












1








1







I'm having an issue inserting php form array values into a MySQL table. Right now the form itself pulls worker information from another table to populate the fields and their values. Right now there are only three workers that would populate those fields, however as more workers get added, there could be as many as 40. When I add just the first worker from the form, all the information inserts normally. However, when I add more than one, the title and employeeId fields are blank and I can't figure out why. Any help would be greatly appreciated.



Here is the form:



<form method = 'POST' action = 'addworkers.php'>
<?php
$sql2 = "select * from workers where companyId = 1";
$result2 = mysqli_query($conn,$sql2);
$numRows = mysqli_num_rows($result2);
$check = 0;
while($row2 = $result2->fetch_assoc()) {
$employeeId = $row2["id"];
$name = $row2["name"];
echo '<input type="checkbox" name="employeeId" value="' . $employeeId . '">';
echo "$namen";
echo '<input type = "hidden" value="' . $companyId . '" name = "companyId"/>';
echo '<input type = "hidden" value="' . $jobNumber . '" name = "jobNumber"/>';
echo 'Site Title : <input type = "text" name = "title"/><br/>';
}
?>
<input type="hidden" name="count" value="<?php echo "$numRows"; ?>"/>
<input type = 'submit' value = 'SEND'/>
</form>


Then the addworkers.php code



require_once("dbConfig.php");
session_start();
$timestamp = date("Y-m-d");

if (isset($_SESSION['loginname'])) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

foreach($title as $key=>$value){
if (!empty($value)) {
$query = "insert into `Jobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId[$key]', '$jobNumber[$key]','$employeeId[$key]','$value','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}


I changed the echo "Error" line but the output was clear.



It must be a problem with how the array is counted but I'm not sure how to fix it. in the form, if I check the boxes next to each row, all the information is entered into the table properly. If I only check the second and/or third line, it doesn't include the Site Title and the employeeId is reversed.



Here is the output of the form:



<form method = 'POST' action = 'insertworkers.php'>
<input type="checkbox" name="employeeId" value="1">Mike
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="2">Steve
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="3">Roger
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="hidden" name="count" value="3"/>
<input type = 'submit' value = 'SEND'/>
</form>


I also changed the foreach loop to a for loop since the array depth should be the same as all fields will be mandatory



require_once("dbConfig.php");
session_start();
$counter = "".$_POST["count"]."";
$timestamp = date("Y-m-d");

if ( isset( $_SESSION['loginname'] ) ) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

for($i=0, $count = count($employeeId);$i<$count;$i++){
if (!empty($employeeId)) {
$query = "insert into `customerJobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId', '$jobNumber','$employeeId[$i]','$title[$i]','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}









share|improve this question















I'm having an issue inserting php form array values into a MySQL table. Right now the form itself pulls worker information from another table to populate the fields and their values. Right now there are only three workers that would populate those fields, however as more workers get added, there could be as many as 40. When I add just the first worker from the form, all the information inserts normally. However, when I add more than one, the title and employeeId fields are blank and I can't figure out why. Any help would be greatly appreciated.



Here is the form:



<form method = 'POST' action = 'addworkers.php'>
<?php
$sql2 = "select * from workers where companyId = 1";
$result2 = mysqli_query($conn,$sql2);
$numRows = mysqli_num_rows($result2);
$check = 0;
while($row2 = $result2->fetch_assoc()) {
$employeeId = $row2["id"];
$name = $row2["name"];
echo '<input type="checkbox" name="employeeId" value="' . $employeeId . '">';
echo "$namen";
echo '<input type = "hidden" value="' . $companyId . '" name = "companyId"/>';
echo '<input type = "hidden" value="' . $jobNumber . '" name = "jobNumber"/>';
echo 'Site Title : <input type = "text" name = "title"/><br/>';
}
?>
<input type="hidden" name="count" value="<?php echo "$numRows"; ?>"/>
<input type = 'submit' value = 'SEND'/>
</form>


Then the addworkers.php code



require_once("dbConfig.php");
session_start();
$timestamp = date("Y-m-d");

if (isset($_SESSION['loginname'])) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

foreach($title as $key=>$value){
if (!empty($value)) {
$query = "insert into `Jobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId[$key]', '$jobNumber[$key]','$employeeId[$key]','$value','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}


I changed the echo "Error" line but the output was clear.



It must be a problem with how the array is counted but I'm not sure how to fix it. in the form, if I check the boxes next to each row, all the information is entered into the table properly. If I only check the second and/or third line, it doesn't include the Site Title and the employeeId is reversed.



Here is the output of the form:



<form method = 'POST' action = 'insertworkers.php'>
<input type="checkbox" name="employeeId" value="1">Mike
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="2">Steve
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="checkbox" name="employeeId" value="3">Roger
<input type = "hidden" value="1" name = "companyId"/>
<input type = "hidden" value="12345" name = "jobNumber"/>
Site Title : <input type = "text" name = "title"/>

<input type="hidden" name="count" value="3"/>
<input type = 'submit' value = 'SEND'/>
</form>


I also changed the foreach loop to a for loop since the array depth should be the same as all fields will be mandatory



require_once("dbConfig.php");
session_start();
$counter = "".$_POST["count"]."";
$timestamp = date("Y-m-d");

if ( isset( $_SESSION['loginname'] ) ) {
$companyId = isset($_POST['companyId']) ? $_POST['companyId'] : "" ;
$jobNumber = isset($_POST['jobNumber']) ? $_POST['jobNumber'] : "" ;
$employeeId = isset($_POST['employeeId']) ? $_POST['employeeId'] : "" ;
$title = isset($_POST['title']) ? $_POST['title'] : "" ;

for($i=0, $count = count($employeeId);$i<$count;$i++){
if (!empty($employeeId)) {
$query = "insert into `customerJobs` (id, companyId, jobId, employeeId, siteTitle, dateAdded) values (NULL,'$companyId', '$jobNumber','$employeeId[$i]','$title[$i]','$timestamp')";
$result = mysqli_query($conn,$query);
}
}
} else {
echo "Error";
}






php mysql arrays forms session






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 at 10:24









DestinatioN

1,2481326




1,2481326










asked Nov 21 at 3:01









Matt

192




192












  • Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
    – Funk Forty Niner
    Nov 21 at 3:03










  • also use error reporting php.net/manual/en/function.error-reporting.php
    – Funk Forty Niner
    Nov 21 at 3:06










  • It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
    – miken32
    Nov 21 at 3:07










  • open to SQL injection attack, dont use this in actual production
    – user10226920
    Nov 21 at 3:10










  • You're only doing a foreach for the one array here, what about the others?
    – Funk Forty Niner
    Nov 21 at 3:11


















  • Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
    – Funk Forty Niner
    Nov 21 at 3:03










  • also use error reporting php.net/manual/en/function.error-reporting.php
    – Funk Forty Niner
    Nov 21 at 3:06










  • It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
    – miken32
    Nov 21 at 3:07










  • open to SQL injection attack, dont use this in actual production
    – user10226920
    Nov 21 at 3:10










  • You're only doing a foreach for the one array here, what about the others?
    – Funk Forty Niner
    Nov 21 at 3:11
















Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
– Funk Forty Niner
Nov 21 at 3:03




Change your echo "Error"; to echo "Error" . mysqli_error($conn); - is there anything?
– Funk Forty Niner
Nov 21 at 3:03












also use error reporting php.net/manual/en/function.error-reporting.php
– Funk Forty Niner
Nov 21 at 3:06




also use error reporting php.net/manual/en/function.error-reporting.php
– Funk Forty Niner
Nov 21 at 3:06












It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
– miken32
Nov 21 at 3:07




It would be more helpful to include the actual HTML of the form, not the PHP that generated it.
– miken32
Nov 21 at 3:07












open to SQL injection attack, dont use this in actual production
– user10226920
Nov 21 at 3:10




open to SQL injection attack, dont use this in actual production
– user10226920
Nov 21 at 3:10












You're only doing a foreach for the one array here, what about the others?
– Funk Forty Niner
Nov 21 at 3:11




You're only doing a foreach for the one array here, what about the others?
– Funk Forty Niner
Nov 21 at 3:11

















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