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Problems displaying value from MySQL database in input field

skateme

By skateme
in PHP Coding Help

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skateme Member

skateme

Posted
Posted

I'm having a lot of trouble displaying a value from a database in one of my input fields. My code is:

 

<?php
session_start();
print "<pre>";print_r($_SESSION);print "</pre>";
?>
<html>
<head>
<title>My Account<title>
</head>
<body>
<?php
$con = mysql_connect("localhost", "admin", "adminpass") or die("Could not connect to database");
$database = mysql_select_db("db") or die("<br />Could not select database");
$username = $_SESSION['username'];
$query=sprintf("select email,location from accountinfo where username='%s'",mysql_real_escape_string($username)); 
$result=mysql_query($query); 
while($row=mysql_fetch_array($result)) 
{ 
$email=$row['email']; 
$location=$row['location']; 
}
?> 
<p>Modify account information:</p>
<form action="modifyaccount.php" method="post">
Username: <input type="text" readonly="readonly" value="<?php echo $username; ?>" /><br />
Password: <input type="password" name="password" id="password" /><br />
Email: <input type="text" name="email" id="email" value="<?php echo $email; ?>" /><br />
Location: <input type="text" name="location" id="location" value="<?php echo $location; ?>" /><br />
<input type="submit" value="Modify" />
</form>
</body>
</html>

 

and the error I get is:

 

Parse error: syntax error, unexpected T_VARIABLE in /Library/WebServer/Documents/MDR/myaccount.php on line 27

 

Line 27 is basically the whole while loop.:

 

while($row = mysql_fetch_array($result))  
{  
$email = $row['email'];  
$location = $row['location'];  
} 

 

What am I doing wrong? Thanks in advance!

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AV1611 Advanced Member

AV1611

Posted
Posted

is there more than one return?

 

while{

suggests there is more than one return.

If{

would be better if you expect only one return.

 

Also,

 

select email,location from accountinfo where username='%s'

is bad.

 

% is a wildcard, and you have to use like not = but you want a single return, so you need a better query...

 

Finally, just for fun do:

));

$result=mysql_query($query) or die('Croaked: '.mysql_error());

 

and see if the query is valid.

 

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JasonLewis Member

JasonLewis

Posted
Posted

Actually using %s is allowed since the OP is using sprintf().

 

But, as you said, why are using a while loop OP when you would surely only be returning 1 row from the database. Just perform mysql_fetch_array() outside the loop then set the variables.

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skateme Member

skateme

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  • Author
Posted

Do you mean something like this:

 

$query  = sprintf("select email from userinfo where username='%s' ", mysql_real_escape_string($username)); 
$locquery  = sprintf("select location from userinfo where username='%s' ", mysql_real_escape_string($location)); 
$result = mysql_query($query);
$locresult = mysql_query($locquery);
$email = mysql_fetch_array($result);
$location = mysql_fetch_array($locresult);

while($result=mysql_query($query))
{ 
$email = $row['email']; 
};
while($locresult=mysql_query($locresult))
{
    $location = $row['location']; 
};

 

I get this error when I use the code above:

 

Parse error: syntax error, unexpected T_VARIABLE in /Library/WebServer/Documents/MDR/myaccount.php on line 25

 

Line 25 is the while loop again.

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JasonLewis Member

JasonLewis

Posted
Posted

Try this:

 

$query = sprintf("SELECT email, location FROM userinfo WHERE username='%s'", mysql_real_escape_string($username)); 
$result = mysql_query($query) or die("Error: ".mysql_error());

$data = mysql_fetch_array($result);

$email = $data['email'];
$location = $data['location'];

echo "Email: " . $email . "<br />Location: " . $location;

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skateme Member

skateme

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  • Author
Posted

Thanks! That worked for the most part. I get an undefined function error:

 

Fatal error: Call to undefined function  sprintf() in /Library/WebServer/Documents/MDR/myaccount.php on line 7

 

My  code:

<?php
session_start(); 
print "<pre>".$_SESSION."</pre>";
$con = mysql_connect("localhost", "admin", "adminpass") or die("Could not connect to database");
$database = mysql_select_db("db") or die("<br />Could not select database");
$username = $_SESSION['username'];
$query  = sprintf("SELECT email, location FROM accountinfo WHERE username = '.$username.'", mysql_real_escape_string($username), $con); 
$result = mysql_query($query); 
$data = mysql_fetch_array($result);

$email = $data['email'];
$location = $data['location'];

?>
<html>
<head>
<title>My Account<title>
</head>
<body>
<p>Modify account information:</p>
<form action="modifyaccount.php" method="post">
Username: <input type="text" readonly="readonly" value="<?php echo $username; ?>" /><br />
Password: <input type="password" name="password" id="password" /><br />
Email: <input type="text" name="email" id="email" value="<?php echo $email; ?>" /><br />
Location: <input type="text" name="location" id="location" value="<?php echo $location; ?>" /><br />
<input type="submit" value="Modify" />
</form>



</body>
</html>

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JasonLewis Member

JasonLewis

Posted
Posted

Hm, well try this then:

 

<?php
session_start(); 
echo "<pre>", print_r($_SESSION), "</pre>";

$con = mysql_connect("localhost", "admin", "adminpass") or die("Could not connect to database");
$database = mysql_select_db("db") or die("<br />Could not select database");
$username = $_SESSION['username'];
$query  = "SELECT email, location FROM accountinfo WHERE username = '".mysql_real_escape_string($username)."'";
$result = mysql_query($query) or die(mysql_error());
$data = mysql_fetch_array($result);

$email = $data['email'];
$location = $data['location'];

?>
<html>
<head>
<title>My Account<title>
</head>
<body>
<p>Modify account information:</p>
<form action="modifyaccount.php" method="post">
Username: <input type="text" readonly="readonly" value="<?php echo $username; ?>" /><br />
Password: <input type="password" name="password" id="password" /><br />
Email: <input type="text" name="email" id="email" value="<?php echo $email; ?>" /><br />
Location: <input type="text" name="location" id="location" value="<?php echo $location; ?>" /><br />
<input type="submit" value="Modify" />
</form>



</body>
</html>

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skateme Member

skateme

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  • Author
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I'm still getting another error message on line 7. Line 7 is the $query line:

 

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in /Library/WebServer/Documents/MDR/myaccount.php on line 7

That error disappears when I remove the quotation marks around SELECT email.., but I get new error saying:

 

Parse error: syntax error, unexpected ',' in /Library/WebServer/Documents/MDR/myaccount.php on line 7

When I remove ",location" from the query, the new error I get is:

 

Parse error: syntax error, unexpected '=' in /Library/WebServer/Documents/MDR/myaccount.php on line 7

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