Trouble with getting data from a MySQL database

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July 2nd, 2009, 12:12 PM

NewWEBdesigner

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PHP5 - Trouble with getting data from a MySQL database

Hello all,

Here is what I am trying to do. A user enters the first name of somebody and that persons first name, last name, and age is displayed. I am using PHP and MySQL. I know the database is set up and working because I can see it in PHP My Admin and I can insert data to it. What is wrong? I think it has something to do with the this part, because this where i edited the tutorial i got the code from:

PHP Code:


		
			
 $result = mysql_query("SELECT * FROM Persons WHERE FirstName='$_POST[firstname]'");

When I try to run this code, I get this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\select.php on line 17

Below is the code I am trying to get to work:

Here is the html code for the form:

Code:

<html> <body>  <form action="select.php" method="post"> Firstname: <input type="text" name="firstname" /> <input type="submit" /> </form>  </body> </html>

Here is the PHP code:

PHP Code:


		
			
<?php   echo "$_POST[firstname]";  $con = mysql_connect("localhost","root",""); if (!$con)   {   die('Could not connect: ' . mysql_error());   }  mysql_select_db("my_db", $con);  $result = mysql_query("SELECT * FROM Persons WHERE FirstName='$_POST[firstname]'");  while($row = mysql_fetch_array($result))   {   echo $row['FirstName'] . " " . $row['LastName'];   echo "<br />";   } ?>

Thanks for your help

July 2nd, 2009, 12:14 PM

ManiacDan

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Do this:

PHP Code:


		
			
 $result = mysql_query("SELECT * FROM Persons WHERE FirstName='" . mysql_real_escape_string($_POST['firstname']) , "'");

if ( !$result ) die(mysql_error());

-Dan

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July 3rd, 2009, 01:03 AM

NewWEBdesigner

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Dan,

Thanks for responding. When I substituted your code for mine I got this error:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\wamp\www\select.php on line 14

here is the code again

PHP Code:


		
			
<?php   echo "$_POST[firstname]";  $con = mysql_connect("localhost","root",""); if (!$con)   {   die('Could not connect: ' . mysql_error());   }  mysql_select_db("test", $con);  $result = mysql_query("SELECT * FROM Persons WHERE FirstName='" . mysql_real_escape_string($_POST['firstname']) , "'"); if ( !$result ) die(mysql_error());    while($row = mysql_fetch_array($result))   {   echo $row['FirstName'] . " " . $row['LastName'];   echo "<br />";   } ?>

July 3rd, 2009, 02:08 AM

DaWizman

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PHP Code:


		
			
<?php

$con = mysql_connect("localhost","root",""); 

if (!$con)   {   

die('Could not connect: ' . mysql_error());   

}  

mysql_select_db("test", $con);

$query = "SELECT `FirstName`, `LastName` FROM `Persons` WHERE `FirstName` = '" . mysql_real_escape_string($_POST['firstname']) . "' ";  

$result = mysql_query($query);

if ( !$result ) {

die(mysql_error());

}    

while($row = mysql_fetch_assoc($result))   {   

echo $row['FirstName'] . " " . $row['LastName'];   

echo "<br />";   

} 

?>

ManiacDan had a comma after the mysql_real_escape_string instead of a period. I also did some housecleaning to make the code a little nicer.

Cheers

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July 10th, 2009, 12:06 AM

NewWEBdesigner

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Thanks guys.

I changed my code around a bit. now I have a database of States and the the colleges in them. I want a user to select the state via a drop down menu. Then when the user goes to select a college, they are presented with only the colleges that correspond to that state.

My problem is that I can't figure out the syntax to get the colleges to appear in a drop down menu without getting errors.

Here is the error message I get:
Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in C:\wamp\www\Drop.php on line 49
(The error refers to the PHP code)

This is my HTML code:

Code:

<html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head>  <body> select your state:   <form action="Drop.php" method="post"> <select name="State"> <option value="NY">NY</option> <option value="ND">ND</option> </select> <input type="submit" /> </form>  <form action="" method=""> <select name="College"> <option value="Select College">Select College</option>  </select> <input type="submit" /> </form> </body> </html>

Here is my php code, drop.php:

PHP Code:


		
			
 Welcome <?php echo $_POST["State"]; ?>!<br />   <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head>  <body> select your state:   <form action="Drop.php" method="post"> <select name="State"> <option value="NY">NY</option> <option value="ND">ND</option> </select> <input type="submit" /> </form>  <?php if (isset($_POST['State'])){     $state = $_POST['State'];  echo "Country Selected: ". $state . "<br />";  }?>   <?php $con = mysql_connect("localhost","root",""); if (!$con)   {   die('Could not connect: ' . mysql_error());   }  mysql_select_db("books", $con);   $result = mysql_query("SELECT * FROM schools WHERE State='$state'"); echo "<form >  <select action=''>";  while($row = mysql_fetch_array($result))   {         echo "<option value='$row['School']'>" . $row['School'] . "</option";      } echo " </select>"; echo " </form>"; ?>  </body> </html>

btw, DaWizman, how did you get my code to appear vertically as opposed to horizontally?

PS: line 49 refers to this line of my code

PHP Code:


		
			
echo "<option value='$row['School']'>" . $row['School'] . "</option";

What's a good online resource for php code syntax? that seems to be what I am having the most trouble with. Thanks

July 10th, 2009, 06:09 AM

IkoTikashi

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PHP Code:


		
			
echo "<option value='$row['School']'>" . $row['School'] . "</option";

should be

PHP Code:


		
			
echo "<option value=\"{$row['School']}\">{$row['School']}</option>";

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July 10th, 2009, 02:38 PM

NewWEBdesigner

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Thank you, it my code works now!!

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