sql query

tr_ellis1973@msn.com · 01:22 AM

I'm getting an error of could not execute the query. not sure what im doing wrong on this query. i'm just messing around with search through a database ofa drink recipe site im building. I wanna use a search option on this website.

PHP Code:


			
<?php

 $host="mysql16.000webhost.com";

 $user ="a7246488_serg91";

 $password = "xxxx";

 $database ="a7246488_drinks";

    $connection = mysql_connect($host, $user, $password) or die ("couldn't connect to server");

    

    

        $db = mysql_select_db($database, $connection) or die("couldn't select database.");

        $name = "drink"; //user types in the name of drink

        

    $query ="select NAME, INGREDIENTS, INSTRUCTIONS, CATEGORY from a1(1) where NAME = '%$name'";

    $result = mysql_query($query) or die ("Couldn't execute the query.");

    

    $row = mysql_num_rows($result);



        //display the result of the table

    $row = mysql_fetch_array($result);

    

    //this is where I think is where im going wrong

    

        echo'Name:'.$row['name'];

        echo'<br>Ingredients: '.$row['Ingredients'];

        echo'<br>instructions: '.$row['Instructions'];

        echo'<br>category: '.$row['category'];

        echo'<br>Image: '.$row['image'];

        

?>

HTML Code:

<body>
<FORM action="search.php" method="post">
Search: <input type="text" name="term" /><br />
<input type="submit" name="submit" value="submit" />
</body>

grace1004 · 02:12 AM

The following code is applicable in case the matching result is just one row.

PHP Code:


			
<?php 

 $host="mysql16.000webhost.com"; 

 $user ="a7246488_serg91"; 

 $password = "xxxx"; 

 $database ="a7246488_drinks"; 



 $connection = mysql_connect($host, $user, $password) or die ("couldn't connect to server");        

 $db = mysql_select_db($database, $connection) or die("couldn't select database."); 



 $name = $_POST['term']; //name of the drink entered by a user

         

 $query ="select NAME, INGREDIENTS, INSTRUCTIONS, CATEGORY from a1(1) where NAME = '%$name%'"; 

 $result = mysql_query($query) or die ("Couldn't execute the query.");     

    

 // display $result in an array

 $row = mysql_fetch_array($result);  

 if($row) { 

        

        echo'Name:'.$row['name']; 

        echo'<br>Ingredients: '.$row['Ingredients']; 

        echo'<br>instructions: '.$row['Instructions']; 

        echo'<br>category: '.$row['category']; 

        echo'<br>Image: '.$row['image'];  

}

else {

echo "No matching result!";

}        

?>

tr_ellis1973@msn.com · 02:33 AM

I double checked what you did. I think its all correct. Still the same error. something has to be killing the select query. I'm just missing it.

PHP Code:


			
$host="mysql16.000webhost.com";

 $user ="a7246488_serg91";

 $password = "xxxx";

 $database ="a7246488_drinks";

    $connection = mysql_connect($host, $user, $password) or die ("couldn't connect to server");

    $db = mysql_select_db($database, $connection) or die("couldn't select database.");

        

    $name = $_POST["NAME"]; //user types in the name of drink

        

    $query = "select NAME, INGREDIENTS, INSTRUCTIONS, CATEGORY from a1(1) where NAME = '%$name%'";

    $result = mysql_query($query) or die ("Couldn't execute the query.");

    

    

//display the result of the table

    $row = mysql_fetch_array($result); 

    if($row)     {

        echo 'Name: <br />'.$row['name'];

        echo 'Ingredients:<br />'.$row['Ingredients'];

        echo 'Instructions:<br /> '.$row['Instructions'];

        echo 'category: <br />'.$row['category'];

        

        

        //echo 'Glass: <br/> '.$row['image'];

    }

else    {

    echo "no matching results!";

}

HTML Code:

<body>
<FORM action="search.php" method="post">
Search: <input type="text" name="NAME" /><br />
<input type="submit" name="submit" value="submit" />
</body>

grace1004 · 04-30-2012, 03:45 AM

Is table name `a1(1)` correct? The table name looks unusual.

tr_ellis1973@msn.com · 04-30-2012, 03:59 AM

yes. I thought it might of been the table name so I changed it to recipes. and im still getting the query failure error.

grace1004 · 04-30-2012, 04:03 AM

Did you check the real table name in your phpmyadmin? Did you change the table name in your phpmyadmin page?

tr_ellis1973@msn.com · 04-30-2012, 04:24 AM

For somew reason its working Know. all I have to do is figure out the lay out and I should be good. Thank you.