MySQL query not working while using php variable in where clause

Question

I'm new to PHP and MySQL. I am trying to make a simple search form using which I would want to show the results from the database based on the input text entered in the form. My code is like this:

Form.php

<!DOCTYPE html>
<html lang="en">
<html>
<head>
<title>test</title>
</head>
<body>
<form action="search.php" method="GET" id="form">
Name: <input type="text" name="name" >
Age:<input type="text" name="age">
Search<input type="submit" name="submit" id="Search" Value="Search">

</form>

</body>
</html>

Connect.php

  <?php
  $connect = mysql_connect('localhost','$user','$password'); 

  if(!$connect){
die('Could not connect'.mysql_error() );  
  }

  $db_selected = mysql_select_db('test');  

  if(!$db_selected){
die('wrong'.mysql_error() );
  }

  ?>

Search.php

 <?php
  include("includes/connect.php");

  $name=$_GET['name'];


  echo $name;

  $query = "SELECT * FROM `cats` WHERE name='\$name'";
  $results= mysql_query($query);


  if (!empty($results)){
echo "query successful" ;
exit;
   }
  $row=mysql_fetch_assoc($results);
  echo "Age:".$row['age'];
  echo "Name:".$row['name'];
 ?>

The echo $names ouputs the result correctly and so does echo "query successful". However the

echo "Age:".$row['age']; 
echo "Name:".$row['name'];

only echo's the string part and the query does not seem to fetch any results.

I tried changing the mysql_fetch_assoc to mysql_fetch_array, but it does not do anything either. Could anyone tell me what am i doing wrong here. My DB table has two columns and two rows.

Also, mysql functions are now deprecated and it's time to move on to PDO or mysqli. — webnoob
– webnoob, Commented Feb 26, 2013 at 19:41

UnholyRanger · Accepted Answer · 2013-02-26 21:03:18Z

3

You're escaping the $ in the variable by doing \$. Try:

$query = "SELECT * FROM `cats` WHERE name='$name'";

EDIT

From the discussion below.

The problem with the undefined index is the fact that you are using $row['age'] when really, the column name in the database is Age. Therefore you must use $row['Age'] when referring to the item. The same goes for name.

edited Feb 26, 2013 at 21:03

answered Feb 26, 2013 at 19:40

UnholyRanger

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15 Comments

user2112593 Over a year ago

Thanks.I tried the query suggested by you. It now gives me "Notice: Undefined index: name and Notice: Undefined index: age " error msg. But if I add qoutes around name ($query = "SELECT * FROM cats WHERE 'name'='$name'";) the error goes away but still the query does not fetch any results.

UnholyRanger Over a year ago

do you have columns named age and name in cats? The quotes are to surround the value and NOT the column name.

user2112593 Over a year ago

Yes, I have a column named age in my table cats.

user2112593 Over a year ago

It gives me " resource(5) of type (mysql result) ".No idea what that means.

UnholyRanger Over a year ago

then try var_dump($row);. mysql_query returns a MySQL resource which you are getting.

|

Christian Rapp · Accepted Answer · 2013-02-26 19:38:56Z

0

$query = "SELECT * FROM `cats` WHERE name='" . $name . "'";

or without concatenation

$query = "SELECT * FROM `cats` WHERE name='$name'";

answered Feb 26, 2013 at 19:38

Christian Rapp

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Comments

Jake Jackson · Accepted Answer · 2013-02-26 19:44:33Z

0

This should work:

$query = "SELECT * FROM cats WHERE name = '$name'";

Also, when you call "exit;" in the following block it will cancel the execution of the rest of your script:

    if (!empty($results)){
     echo "query successful" ;
     exit;
    }

answered Feb 26, 2013 at 19:44

Jake Jackson

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