Issue with index - unidentified index [duplicate]
-2
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
28 answers
PHP ereg_replace deprecated
2 answers
Warning: mysqli_error() expects exactly 1 parameter, 0 given error [closed]
3 answers
The issue appears to be with an "unidentified index - pid". What I am trying to do is to show the body text of a linked page but get the error mentioned. I have followed a working example online and the code seems the same.
This is the php:
//Determine which page to use
if (!$_GET['pid']) {
$pageid = '1';
} else {
$pageid = ereg_replace("[^0-9]", "", $_GET['pid']); // filter everything but numbers for security
}
// Query the body section for the proper page
$sqlCommand = "SELECT pagebody FROM pages WHERE id='$pageid' LIMIT 1";
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error());
while ($row = mysqli_fetch_array($query)) {
$body = $row["pagebody"];
}
mysqli_free_result($query);
// end body query
I had noticed that in the example, the working code was placed on an online server, whereas I am using XAMPP. Could this be the issue? Will the code only work online?
php mysql indexing get
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
marked as duplicate by aynber, Funk Forty Niner php
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Nov 20 at 19:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
-2
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
28 answers
PHP ereg_replace deprecated
2 answers
Warning: mysqli_error() expects exactly 1 parameter, 0 given error [closed]
3 answers
The issue appears to be with an "unidentified index - pid". What I am trying to do is to show the body text of a linked page but get the error mentioned. I have followed a working example online and the code seems the same.
This is the php:
//Determine which page to use
if (!$_GET['pid']) {
$pageid = '1';
} else {
$pageid = ereg_replace("[^0-9]", "", $_GET['pid']); // filter everything but numbers for security
}
// Query the body section for the proper page
$sqlCommand = "SELECT pagebody FROM pages WHERE id='$pageid' LIMIT 1";
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error());
while ($row = mysqli_fetch_array($query)) {
$body = $row["pagebody"];
}
mysqli_free_result($query);
// end body query
I had noticed that in the example, the working code was placed on an online server, whereas I am using XAMPP. Could this be the issue? Will the code only work online?
php mysql indexing get
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
marked as duplicate by aynber, Funk Forty Niner php
Users with the php badge can single-handedly close php questions as duplicates and reopen them as needed.
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Nov 20 at 19:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It's better to use!issetoremptyinstead of just!
– aynber
Nov 20 at 18:56
You can better check!$_GET['pid']with!isset($_GET['pid']). On which line do you get this error and what is that line?
– vivek_23
Nov 20 at 18:58
You might also like to check the 2nd duplicate, just in case. Thatereg_replace()is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php
– Funk Forty Niner
Nov 20 at 19:07
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10
add a comment |
-2
-2
-2
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
28 answers
PHP ereg_replace deprecated
2 answers
Warning: mysqli_error() expects exactly 1 parameter, 0 given error [closed]
3 answers
The issue appears to be with an "unidentified index - pid". What I am trying to do is to show the body text of a linked page but get the error mentioned. I have followed a working example online and the code seems the same.
This is the php:
//Determine which page to use
if (!$_GET['pid']) {
$pageid = '1';
} else {
$pageid = ereg_replace("[^0-9]", "", $_GET['pid']); // filter everything but numbers for security
}
// Query the body section for the proper page
$sqlCommand = "SELECT pagebody FROM pages WHERE id='$pageid' LIMIT 1";
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error());
while ($row = mysqli_fetch_array($query)) {
$body = $row["pagebody"];
}
mysqli_free_result($query);
// end body query
I had noticed that in the example, the working code was placed on an online server, whereas I am using XAMPP. Could this be the issue? Will the code only work online?
php mysql indexing get
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
28 answers
PHP ereg_replace deprecated
2 answers
Warning: mysqli_error() expects exactly 1 parameter, 0 given error [closed]
3 answers
The issue appears to be with an "unidentified index - pid". What I am trying to do is to show the body text of a linked page but get the error mentioned. I have followed a working example online and the code seems the same.
This is the php:
//Determine which page to use
if (!$_GET['pid']) {
$pageid = '1';
} else {
$pageid = ereg_replace("[^0-9]", "", $_GET['pid']); // filter everything but numbers for security
}
// Query the body section for the proper page
$sqlCommand = "SELECT pagebody FROM pages WHERE id='$pageid' LIMIT 1";
$query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error());
while ($row = mysqli_fetch_array($query)) {
$body = $row["pagebody"];
}
mysqli_free_result($query);
// end body query
I had noticed that in the example, the working code was placed on an online server, whereas I am using XAMPP. Could this be the issue? Will the code only work online?
This question already has an answer here:
“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
28 answers
PHP ereg_replace deprecated
2 answers
Warning: mysqli_error() expects exactly 1 parameter, 0 given error [closed]
3 answers
php mysql indexing get
php mysql indexing get
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
edited Nov 20 at 19:10
Funk Forty Niner
80.5k1247100
80.5k1247100
asked Nov 20 at 18:53
user3511057
388
asked Nov 20 at 18:53
user3511057
388
asked Nov 20 at 18:53
user3511057
388
388
marked as duplicate by aynber, Funk Forty Niner php
Users with the php badge can single-handedly close php questions as duplicates and reopen them as needed.
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Nov 20 at 19:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by aynber, Funk Forty Niner php
Users with the php badge can single-handedly close php questions as duplicates and reopen them as needed.
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Nov 20 at 19:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It's better to use!issetoremptyinstead of just!
– aynber
Nov 20 at 18:56
You can better check!$_GET['pid']with!isset($_GET['pid']). On which line do you get this error and what is that line?
– vivek_23
Nov 20 at 18:58
You might also like to check the 2nd duplicate, just in case. Thatereg_replace()is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php
– Funk Forty Niner
Nov 20 at 19:07
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10
add a comment |
It's better to use!issetoremptyinstead of just!
– aynber
Nov 20 at 18:56
You can better check!$_GET['pid']with!isset($_GET['pid']). On which line do you get this error and what is that line?
– vivek_23
Nov 20 at 18:58
You might also like to check the 2nd duplicate, just in case. Thatereg_replace()is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php
– Funk Forty Niner
Nov 20 at 19:07
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10
It's better to use
!isset or empty instead of just !– aynber
Nov 20 at 18:56
It's better to use
!isset or empty instead of just !– aynber
Nov 20 at 18:56
You can better check
!$_GET['pid'] with !isset($_GET['pid']). On which line do you get this error and what is that line?– vivek_23
Nov 20 at 18:58
You can better check
!$_GET['pid'] with !isset($_GET['pid']). On which line do you get this error and what is that line?– vivek_23
Nov 20 at 18:58
You might also like to check the 2nd duplicate, just in case. That
ereg_replace() is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php– Funk Forty Niner
Nov 20 at 19:07
You might also like to check the 2nd duplicate, just in case. That
ereg_replace() is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php– Funk Forty Niner
Nov 20 at 19:07
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10
add a comment |
1 Answer
1
active
oldest
votes
0
before accessing $_GET['pid'] check if pid set or not
if (empty($_GET['pid'])) {
$pageid = '1';
} else {
.....
}
answered Nov 20 at 18:57
suresh bambhaniya
861113
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
Would then be quicker to writeif(empty($_GET['pid']), that'sisset()and!in one go.
– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
before accessing $_GET['pid'] check if pid set or not
if (empty($_GET['pid'])) {
$pageid = '1';
} else {
.....
}
answered Nov 20 at 18:57
suresh bambhaniya
861113
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
Would then be quicker to writeif(empty($_GET['pid']), that'sisset()and!in one go.
– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
add a comment |
0
before accessing $_GET['pid'] check if pid set or not
if (empty($_GET['pid'])) {
$pageid = '1';
} else {
.....
}
answered Nov 20 at 18:57
suresh bambhaniya
861113
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
Would then be quicker to writeif(empty($_GET['pid']), that'sisset()and!in one go.
– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
add a comment |
0
0
0
before accessing $_GET['pid'] check if pid set or not
if (empty($_GET['pid'])) {
$pageid = '1';
} else {
.....
}
answered Nov 20 at 18:57
suresh bambhaniya
861113
before accessing $_GET['pid'] check if pid set or not
if (empty($_GET['pid'])) {
$pageid = '1';
} else {
.....
}
answered Nov 20 at 18:57
suresh bambhaniya
861113
edited Nov 20 at 19:10
answered Nov 20 at 18:57
suresh bambhaniya
861113
answered Nov 20 at 18:57
suresh bambhaniya
861113
answered Nov 20 at 18:57
suresh bambhaniya
861113
861113
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
Would then be quicker to writeif(empty($_GET['pid']), that'sisset()and!in one go.
– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
add a comment |
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
Would then be quicker to writeif(empty($_GET['pid']), that'sisset()and!in one go.
– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
1
1
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
Thats a redundant check. Only The isset check is needed.
– Ray
Nov 20 at 19:00
1
1
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
i thought !$_GET['pid'] is for 0
– suresh bambhaniya
Nov 20 at 19:02
1
1
Would then be quicker to write
if(empty($_GET['pid']), that's isset() and ! in one go.– minitauros
Nov 20 at 19:05
Would then be quicker to write
if(empty($_GET['pid']), that's isset() and ! in one go.– minitauros
Nov 20 at 19:05
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
This is a 50% fix.
– Funk Forty Niner
Nov 20 at 19:08
add a comment |
It's better to use
!issetoremptyinstead of just!– aynber
Nov 20 at 18:56
You can better check
!$_GET['pid']with!isset($_GET['pid']). On which line do you get this error and what is that line?– vivek_23
Nov 20 at 18:58
You might also like to check the 2nd duplicate, just in case. That
ereg_replace()is deprecated and is most likely contributing to the error. Enable error reporting php.net/manual/en/function.error-reporting.php– Funk Forty Niner
Nov 20 at 19:07
You are leaving yourself open to an sql injection; use a prepared statement.
– Funk Forty Niner
Nov 20 at 19:10