Mysql switch statement doesnt work right...

[Russia]

Hey guys, I have a mysql switch statement that shows different things based on the row value.

 

Here it is...

 

<?php
switch ($rows['icon']) {
    case 1:
        $picture = '<img src=\"img/apple.gif\" title=\"apple\" alt=\"apple\" />';
        echo $picture;
        break;
        
    case 2:
        $picture = '<img src=\"img/banana.gif\" title=\"banana\" alt=\"banana\" />';
        echo $picture;
        break;
        
    case 3:
        $picture = '<img src=\"img/orange.gif\" title=\"orange\" alt=\"orange\" />';
        echo $picture;
        break;
        
    default:
        echo "$rows[icon] is something other than 1 2 or 3";
        break;
}  
?>

 

 

Here is how the other code looks like.

 

 

<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');


mysql_connect("localhost", "", "")or die("cannot connect");
mysql_select_db("test")or die("cannot select DB");
$tbl_name="test_mysql";
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
$count=mysql_num_rows($result);

if (isset($_POST['Submit'])) {
    for($i=0;$i<$count;$i++){
        $month   = $_POST['month'];
        $date    = $_POST['date'];
        $message = $_POST['message'];
        $title   = $_POST['title'];
        $id      = $_POST['id'];
        $icon    = $_POST['icon'];         // ILYA
        $monthday= $month[$i]."<br>".$date[$i];
        $sql1="UPDATE $tbl_name 
                    SET monthday='$monthday', 
                        month='$month[$i]', 
                        date='$date[$i]', 
                        message='" . mysql_real_escape_string($message[$i]) . "', 
                        title='" . mysql_real_escape_string($title[$i]) . "',
                        icon='$icon[$i]'                        
                    WHERE id=".$id[$i];                        // ILYA

    if(!($result1 = mysql_query($sql1))){                   
        "<BR>Error UPDATING $tbl_name ";                              
        exit();                                             
        }                                                   
    }
}
$result=mysql_query($sql);
$count=mysql_num_rows($result);


?>

 

Dont mind the other UPDATE code its part of something else.

 

Anyways, why is the case automaticaly going to the last one... with the words: is something other than 1 2 or 3

 

 

Also, this is with multiple rows at once so I think that that may the problem...

[Pikachu2000]

The value of $rows['icon'] must equal something other than 1, 2 or 3. Have you tried echoing it to see what it actually is? Nothing in the second code chunk assigns a value to $rows['icon'], BTW.

[Russia]

Yeah here is a picture of my database.

 

 

Can you help me fix this issue?

[Russia]

Yeah I have no idea on how to make this work with multiple rows...

[Russia]

Somone told me to do this "set $picture to an array and when you call on the picture just put the number you want (IE 5 items - $picture[0] thru $picture[4])"

 

I really dont understand how to do that.

[Russia]

bump. please help.

[Pikachu2000]

Do not bump threads.

[Russia]

Sorry mate, but i really am not understanding this. Can you help me out? not being rude mate

[Russia]

Okay I think I for part of it to work:

<?php

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

switch ($result['icon']) {
    case 1:
        $picture = '<img src=\"img/apple.gif\" title=\"apple\" alt=\"apple\" />';
        echo $picture;
        break;
        
    case 2:
        $picture = '<img src=\"img/banana.gif\" title=\"banana\" alt=\"banana\" />';
        echo $picture;
        break;
        
    case 3:
        $picture = '<img src=\"img/orange.gif\" title=\"orange\" alt=\"orange\" />';
        echo $picture;
        break;
        
    default:
        echo "$row[icon] is something other than 1 2 or 3";
        break;
}   



}  
?>

 

The thing is it still says

 

1 is something other than 1 2 or 3

2 is something other than 1 2 or 3

3 is something other than 1 2 or 3

3 is something other than 1 2 or 3

2 is something other than 1 2 or 3

1 is something other than 1 2 or 3

 

I need it to show the image.

 

The thing is the values in the rows are 1, 2, 3, 3, 2, 1.

 

In that order from the first row to the last.

[AbraCadaver]

switch ($row['icon']) {

 

Not:

 

switch ($result['icon']) {

[AbraCadaver]

This will tell you that:

 

error_reporting(E_ALL);
ini_set('display_errors', '1');

[Russia]

Geat. Now how can I echo the values around the website. Not exactly in order, like on the bottom of the page I want to echo like row 2's image BY ITSELF!

 

and another part I want to echo like the image from row 5. How would I exactly do that?

 

Basiclly I dont want it all in order.

 

 

Should i use something like:

 

<php?
echo $picture['2']
?>

 

The thing in the brackets [] would be the row i want to echo from...

Is that at all possible?

[AbraCadaver]

$picture is not an array.  It is assigned exactly one value in the switch statement and is overwritten each time through the loop.  You could do $picture[$id] to create an array of every row:

 

while($row = mysql_fetch_assoc($result))
{
$id = $row['id'];

switch ($row['icon']) {
    case 1:
        $picture[$id] = '<img src=\"img/apple.gif\" title=\"apple\" alt=\"apple\" />';
        echo $picture[$id];
        break;
        
    case 2:
        $picture[$id] = '<img src=\"img/banana.gif\" title=\"banana\" alt=\"banana\" />';
        echo $picture[$id];
        break;
        
    case 3:
        $picture[$id] = '<img src=\"img/orange.gif\" title=\"orange\" alt=\"orange\" />';
        echo $picture[$id];
        break;
        
    default:
        $picture[$id] = '';
        echo $row['icon'] . " is something other than 1 2 or 3";
        break;
}   

 

Then echo $picture[1] would echo the picture from the row with id = 1.

[Russia]

Oh my god thank you ssssoooo much

 

Also, for some reason im having a problem of the form not showing up...

 

When I load the page it gives me this error:

 

The lines of code its reffering too is:

<script language="javascript" type="text/javascript">
function showvalue(arg) {
alert(arg);
//arg.visible(false);
}
$(document).ready(function() {
try {
	oHandler = $(".mydds").msDropDown().data("dd");
	oHandler.visible(true);
	//alert($.msDropDown.version);
	//$.msDropDown.create("body select");
	$("#ver").html($.msDropDown.version);
	} catch(e) {
	alert("Error: "+e.message);
	}
})	
</script>

 

The thing is, when I remove that code that we just worked on together the problem goes away... why is this happening?

 

 

[AbraCadaver]

Needs to go in the javascript forum (looks like jquery?).

[Russia]

Yeah, I was just wondering if you really had any idea haha. Okay il post it there.

 

The thing is, what should i post? my whole code, or just the part we worked on?

[Russia]

Okay sorry I managed to fix that part, i moved it below some other code and it now gives me an error like this:

Notice: Undefined variable: picture in C:\wamp\www\updat5.php on line 145

 

This is line 145:

echo $picture[2];

 

and its part of this code:

 

<?php

while($row = mysql_fetch_assoc($result))
{
$id = $row['id'];

switch ($row['icon']) {
    case 1:
        $picture[$id] = '<img src="img/apple.gif" title="apple" alt="apple" />';
        echo $picture[$id];
        break;
        
    case 2:
        $picture[$id] = '<img src="img/banana.gif" title="banana" alt="banana" />';
        echo $picture[$id];
        break;
        
    case 3:
        $picture[$id] = '<img src="img/orange.gif" title="orange" alt="orange" />';
        echo $picture[$id];
        break;
        
    default:
        $picture[$id] = '';
        echo $row['icon'] . " is something other than 1 2 or 3";
        break;
}  
}


<?php
echo $picture[2];
?>

 

[Russia]

Okay, I posted it outside the while and now it gives me a new error:

 

Notice: Undefined offset: 2 in C:\wamp\www\updat5.php on line 147

 

<?php
$picture = array();
while($row = mysql_fetch_assoc($result))
{
$id = $row['id'];
$picture = array();
switch ($row['icon']) {
    case 1:
        $picture[$id] = '<img src="img/apple.gif" title="apple" alt="apple" />';
        echo $picture[$id];
        break;
        
    case 2:
        $picture[$id] = '<img src="img/banana.gif" title="banana" alt="banana" />';
        echo $picture[$id];
        break;
        
    case 3:
      $picture[$id] = '<img src="img/orange.gif" title="orange" alt="orange" />';
        echo $picture[$id];
        break;
        
    default:
        $picture[$id] = '';
        echo $row['icon'] . " is something other than 1 2 or 3";
        break;
}  
}
?>

<?php
echo $picture[2];
?>

 

My next post will be the full code if needed incase the error I have currently cant be fixed with the current snippet/block i posted.