Select data from a row in database
I am making a profile page in which I want the user's email to be displayed. I thought this would be quite a simple code that could be achieved using the select function from the database. However, this only works for one string and I cannot seem to figure out why.
This is my original code
session_start();
$_SESSION["user"] = $username;
$_SESSION["pass"] = $password;
$_SESSION["email"] = $email;
$connection = mysqli_connect ("localhost", "root", "", "picshare");
if ($connection ->connect_error) {
die("Connection failed: " . $connection->connect_error);
}else{
$query = mysqli_query($connection, "SELECT email FROM login WHERE username='".$_SESSION["user"]."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
}else{
echo "error";
$conn->close();
}}
When I try and echo $field, nothing was echoed
<p class ="right uc"><?php echo($field['email']);?></p>
I retried the code, but instead of using a session, I made a variable
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
$query = mysqli_query($conn, "SELECT email FROM login WHERE username='".$user."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
echo($field['email']);
}else{
echo "error";
$conn->close();
This code displayed the email, and so I proceeded to change the $user variable to another name and the original problem occurred where nothing was echoed.
I went back to the original code and I logged in as Eniola Olaogun and the email was echoed, but as soon as I changed the person I logged in as, no email was echoed.
I am not sure why I am experiencing this problem and some help would be greatly appreciated
php html mysql database session
add a comment |
I am making a profile page in which I want the user's email to be displayed. I thought this would be quite a simple code that could be achieved using the select function from the database. However, this only works for one string and I cannot seem to figure out why.
This is my original code
session_start();
$_SESSION["user"] = $username;
$_SESSION["pass"] = $password;
$_SESSION["email"] = $email;
$connection = mysqli_connect ("localhost", "root", "", "picshare");
if ($connection ->connect_error) {
die("Connection failed: " . $connection->connect_error);
}else{
$query = mysqli_query($connection, "SELECT email FROM login WHERE username='".$_SESSION["user"]."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
}else{
echo "error";
$conn->close();
}}
When I try and echo $field, nothing was echoed
<p class ="right uc"><?php echo($field['email']);?></p>
I retried the code, but instead of using a session, I made a variable
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
$query = mysqli_query($conn, "SELECT email FROM login WHERE username='".$user."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
echo($field['email']);
}else{
echo "error";
$conn->close();
This code displayed the email, and so I proceeded to change the $user variable to another name and the original problem occurred where nothing was echoed.
I went back to the original code and I logged in as Eniola Olaogun and the email was echoed, but as soon as I changed the person I logged in as, no email was echoed.
I am not sure why I am experiencing this problem and some help would be greatly appreciated
php html mysql database session
1
mysqli_error($con)
that won't work, you're using a different variable.
– Funk Forty Niner
Nov 24 '18 at 17:30
add a comment |
I am making a profile page in which I want the user's email to be displayed. I thought this would be quite a simple code that could be achieved using the select function from the database. However, this only works for one string and I cannot seem to figure out why.
This is my original code
session_start();
$_SESSION["user"] = $username;
$_SESSION["pass"] = $password;
$_SESSION["email"] = $email;
$connection = mysqli_connect ("localhost", "root", "", "picshare");
if ($connection ->connect_error) {
die("Connection failed: " . $connection->connect_error);
}else{
$query = mysqli_query($connection, "SELECT email FROM login WHERE username='".$_SESSION["user"]."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
}else{
echo "error";
$conn->close();
}}
When I try and echo $field, nothing was echoed
<p class ="right uc"><?php echo($field['email']);?></p>
I retried the code, but instead of using a session, I made a variable
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
$query = mysqli_query($conn, "SELECT email FROM login WHERE username='".$user."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
echo($field['email']);
}else{
echo "error";
$conn->close();
This code displayed the email, and so I proceeded to change the $user variable to another name and the original problem occurred where nothing was echoed.
I went back to the original code and I logged in as Eniola Olaogun and the email was echoed, but as soon as I changed the person I logged in as, no email was echoed.
I am not sure why I am experiencing this problem and some help would be greatly appreciated
php html mysql database session
I am making a profile page in which I want the user's email to be displayed. I thought this would be quite a simple code that could be achieved using the select function from the database. However, this only works for one string and I cannot seem to figure out why.
This is my original code
session_start();
$_SESSION["user"] = $username;
$_SESSION["pass"] = $password;
$_SESSION["email"] = $email;
$connection = mysqli_connect ("localhost", "root", "", "picshare");
if ($connection ->connect_error) {
die("Connection failed: " . $connection->connect_error);
}else{
$query = mysqli_query($connection, "SELECT email FROM login WHERE username='".$_SESSION["user"]."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
}else{
echo "error";
$conn->close();
}}
When I try and echo $field, nothing was echoed
<p class ="right uc"><?php echo($field['email']);?></p>
I retried the code, but instead of using a session, I made a variable
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}else{
$query = mysqli_query($conn, "SELECT email FROM login WHERE username='".$user."'");
$field = mysqli_fetch_assoc($query);
if (!$query)
{
die('Error: ' . mysqli_error($con));
}if(mysqli_num_rows($query) > 0){
$field = mysqli_fetch_assoc($query);
echo($field['email']);
}else{
echo "error";
$conn->close();
This code displayed the email, and so I proceeded to change the $user variable to another name and the original problem occurred where nothing was echoed.
I went back to the original code and I logged in as Eniola Olaogun and the email was echoed, but as soon as I changed the person I logged in as, no email was echoed.
I am not sure why I am experiencing this problem and some help would be greatly appreciated
php html mysql database session
php html mysql database session
edited Nov 24 '18 at 17:30
Funk Forty Niner
1
1
asked Nov 24 '18 at 16:20
E. OlaogunE. Olaogun
1
1
1
mysqli_error($con)
that won't work, you're using a different variable.
– Funk Forty Niner
Nov 24 '18 at 17:30
add a comment |
1
mysqli_error($con)
that won't work, you're using a different variable.
– Funk Forty Niner
Nov 24 '18 at 17:30
1
1
mysqli_error($con)
that won't work, you're using a different variable.– Funk Forty Niner
Nov 24 '18 at 17:30
mysqli_error($con)
that won't work, you're using a different variable.– Funk Forty Niner
Nov 24 '18 at 17:30
add a comment |
2 Answers
2
active
oldest
votes
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "SELECT email FROM login WHERE username= {$user}";
$result = $conn->query($sql);
if($result->num_rows > 0) {
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo($row['email']);
}
else {
echo "error";
$conn->close();
add a comment |
error connecting to mysql database through crazy domains
It sounds like it is a permission issue based on the user you're logging in as.
Test it with a 'root' user password that has global access and then troubleshoot and isolate it from there. I'm betting you will find it then.
Pretty much impossible for me to test this remotely since I don't have your DB schema and user accounts to validate with.
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53460075%2fselect-data-from-a-row-in-database%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "SELECT email FROM login WHERE username= {$user}";
$result = $conn->query($sql);
if($result->num_rows > 0) {
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo($row['email']);
}
else {
echo "error";
$conn->close();
add a comment |
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "SELECT email FROM login WHERE username= {$user}";
$result = $conn->query($sql);
if($result->num_rows > 0) {
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo($row['email']);
}
else {
echo "error";
$conn->close();
add a comment |
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "SELECT email FROM login WHERE username= {$user}";
$result = $conn->query($sql);
if($result->num_rows > 0) {
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo($row['email']);
}
else {
echo "error";
$conn->close();
$host = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "picshare";
$user = 'Eniola Olaogun';
$conn = new mysqli ($host, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else {
$sql = "SELECT email FROM login WHERE username= {$user}";
$result = $conn->query($sql);
if($result->num_rows > 0) {
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo($row['email']);
}
else {
echo "error";
$conn->close();
answered Nov 25 '18 at 4:00
Mohsin MujawarMohsin Mujawar
666
666
add a comment |
add a comment |
error connecting to mysql database through crazy domains
It sounds like it is a permission issue based on the user you're logging in as.
Test it with a 'root' user password that has global access and then troubleshoot and isolate it from there. I'm betting you will find it then.
Pretty much impossible for me to test this remotely since I don't have your DB schema and user accounts to validate with.
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
add a comment |
error connecting to mysql database through crazy domains
It sounds like it is a permission issue based on the user you're logging in as.
Test it with a 'root' user password that has global access and then troubleshoot and isolate it from there. I'm betting you will find it then.
Pretty much impossible for me to test this remotely since I don't have your DB schema and user accounts to validate with.
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
add a comment |
error connecting to mysql database through crazy domains
It sounds like it is a permission issue based on the user you're logging in as.
Test it with a 'root' user password that has global access and then troubleshoot and isolate it from there. I'm betting you will find it then.
Pretty much impossible for me to test this remotely since I don't have your DB schema and user accounts to validate with.
error connecting to mysql database through crazy domains
It sounds like it is a permission issue based on the user you're logging in as.
Test it with a 'root' user password that has global access and then troubleshoot and isolate it from there. I'm betting you will find it then.
Pretty much impossible for me to test this remotely since I don't have your DB schema and user accounts to validate with.
answered Nov 24 '18 at 16:33
WaxhawWaxhaw
274
274
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
add a comment |
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
check code there is root user for connection already.
– Akhilesh
Nov 24 '18 at 17:21
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53460075%2fselect-data-from-a-row-in-database%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
mysqli_error($con)
that won't work, you're using a different variable.– Funk Forty Niner
Nov 24 '18 at 17:30