Member Avatar for jkulp4

Hi All,

I am a newcomer to PHP and still learning. Any help with this problem would be appreciated.
Please see the code below. Whenever I run this I get an error message:

"Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in
C:\wamp\www\selected.php on line 25"

Can someone tell me what Iam doing wrong?

<html>
<Head>
<Title>Drop Down Choice Example</title>

<?php

//connect to database, checking, etc
include_once "connect.php";

// process form when posted
if(isset($_POST['value'])) {
    if($_POST['value'] == 'Kulp') {
        // query to get all Kulp records  
        $query = "SELECT * FROM names WHERE LastName='Kulp'";  
    }  
    elseif($_POST['value'] == 'Smith') {  
        // query to get all Smith records  
        $query = "SELECT * FROM names WHERE LastName='Smith'";  
    } else {  
        // query to get all records  
        $query = "SELECT * FROM names";  
    }  
    $sql = mysql_query($query);  

    while ($row = mysql_fetch_array($query)){ 
        $Id = $row["Id"]; 
        $FirstName = $row["FirstName"]; 
        $LastName = $row["LastName"]; 

        // Echo your rows here... 
        echo 'The user ID is:' . $row['id'];
    }
 }
?>
</head>

<body>
<P>Please select from below<p><BR>

<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' >
        <select name="value">
        <option value="All">All</option>
        <option value="Kulp">Kulp</option>
        <option value="Smith">Smith</option>
    </select>
    <br />
    <input type='submit' value = 'Filter'>
</form>
  • 2 Contributors
  • 5 Replies
  • 124 Views
  • 4 Hours Discussion Span
  • Latest Post Latest Post by phorce
Member Avatar for phorce

Hey, please try the following:

<html>
<Head>
<Title>Drop Down Choice Example</title>

<?php

//connect to database, checking, etc
include_once "connect.php";

// process form when posted
if(isset($_POST['value'])) {
    if($_POST['value'] == 'Kulp') {
        // query to get all Kulp records  
        $query = "SELECT * FROM names WHERE LastName='Kulp'";  
    }  
    elseif($_POST['value'] == 'Smith') {  
        // query to get all Smith records  
        $query = "SELECT * FROM names WHERE LastName='Smith'";  
    } else {  
        // query to get all records  
        $query = "SELECT * FROM names";  
    }  
    $sql = mysql_query($query);  

    while ($row = mysql_fetch_array($sql)){ 
        $Id = $row["Id"]; 
        $FirstName = $row["FirstName"]; 
        $LastName = $row["LastName"]; 

        // Echo your rows here... 
        echo 'The user ID is:' . $row['id'];
    }
 }
?>
</head>

<body>
<P>Please select from below<p><BR>

<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' >
        <select name="value">
        <option value="All">All</option>
        <option value="Kulp">Kulp</option>
        <option value="Smith">Smith</option>
    </select>
    <br />
    <input type='submit' value = 'Filter'>
</form>

Your problem is that $query just stores what the query should perform, and, $sql is what executes the query, therefore, you need to pass in this value rather than the string.

Hope this solves your problem

Member Avatar for jkulp4

phorce,

Thanks for the reply mate - that makes sense to me. I knew I was overlooking something.
I changed the code to reflect

while ($row = mysql_fetch_array($sql)){

However, I am still reciving the error message.
Any additional thoughts?

Member Avatar for phorce

I don't get an error message doing this:

<?php

//connect to database, checking, etc
include_once "connect.php";

// process form when posted
if(isset($_POST['value'])) {


    if($_POST['value'] == 'Kulp') {
        // query to get all Kulp records  
        $query = "SELECT * FROM names WHERE LastName='Kulp'"; 
        $sql = mysql_query($query);  
    }  
    elseif($_POST['value'] == 'Smith') {  
        // query to get all Smith records  
        $query = "SELECT * FROM names WHERE LastName='Smith'";  
        $sql = mysql_query($query); 
    } else {  
        // query to get all records  
        $query = "SELECT * FROM names"; 
        $sql = mysql_query($query);  
    }  
    while ($row = mysql_fetch_array($sql)){ 
        $Id = $row["Id"]; 
        $FirstName = $row["FirstName"]; 
        $LastName = $row["LastName"]; 

        // Echo your rows here... 
        echo 'The user ID is:' . $row['id'];
    }
 }
?>

I don't have your server, so I can't really test it.. But give it a try..

commented: Quick reply and very helpful +0
Member Avatar for jkulp4

Phorce,

Thank you! I am an idiot sometimes - had the wrong sql table name in my script.
Your solution works perfectly. Thanks so much for the help. Time for my to take a break for the night obviously my brain is fried. :-)
Thanks again. Much appreciated.

Member Avatar for phorce

No problem mate :) Good luck with it! Please mark this thread as solved, if you haven't done so already! Get some sleep haha

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