Database if else

[forumnz]

Hello there,

 

The code below is in a document that is meant to display one image if it == 0 and another if == 1. It works ok but the problem is that the first image in the code shows up no matter what and then the changing image next to it.

 

What do I do?

 

<?php

$database=mysql_connect("localhost","name", "pass");

mysql_select_db("test",$database);

$query="select * from members";

$result=mysql_query($query);

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

$image_num=$img['stg1'];

if($image_num==0){

echo"<img src='images/not_completed.gif'></img>";

}elseif($image_num==1){

echo"<img src='images/completed.gif'></img>";

}else {

echo" sorry there is no images to show";

}
}
?>

[Jessica]

um, do you have one result that comes to 0 and another that comes to 1? You're selecting more than one row...

[forumnz]

I have a row named stg1 and it can either be set to 1 or 0. Make sense?

[forumnz]

Anyone please?

[JJohnsenDK]

Well i guess that you write either 0 or 1 in the database which would be strings so you also have to refer to strings, like this:

 

<?php

$database=mysql_connect("localhost","name", "pass");

mysql_select_db("test",$database);

$query="select * from members";

$result=mysql_query($query);

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

$image_num=$img['stg1'];

if($image_num=="0"){

echo"<img src='images/not_completed.gif'></img>";

}elseif($image_num=="1"){

echo"<img src='images/completed.gif'></img>";

}else {

echo" sorry there is no images to show";

}
}
?>

 

You also could use NULL then you would have to do something like this:

 

if($image_num==NULL){

echo"<img src='images/not_completed.gif'></img>";

}elseif($image_num!=NULL){

echo"<img src='images/completed.gif'></img>";

}else {

[Jessica]

Yes, and if you have one entry in your DB which has a 0, and one which has a 1, you'll get the results you're getting.

 

JJ: PHP uses weak typing so 0 == "0" == '0' == false.

[forumnz]

It still comes up with multiple images?

[Jessica]

What are the entries in your database?

By the way, there is no </img> tag - img is one tag.

<img src="image.gif" />

[JJohnsenDK]

aaah... allright thanks... didnt know that, nice to know  ...

[forumnz]

This is what I have - also, is there any way to get different results for each user based on their id?

 

-- 
-- Table structure for table `members`
-- 

CREATE TABLE `members` (
  `id` int(4) NOT NULL auto_increment,
  `username` varchar(45) default NULL,
  `password` varchar(45) default NULL,
  `active` int(11) NOT NULL default '0',
  `stg1` int(1) NOT NULL default '1',
  `stg2` int(1) NOT NULL default '0',
  `stg3` int(1) NOT NULL default '0',
  `stg4` int(1) NOT NULL default '0',
  `stg5` int(1) NOT NULL default '0',
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;

-- 
-- Dumping data for table `members`
-- 

INSERT INTO `members` VALUES (1, 'admin', 'pass', 0, 0, 0, 0, 0, 0);
INSERT INTO `members` VALUES (2, 'admin1234', 'pass', 1, 1, 0, 1, 0, 1);
INSERT INTO `members` VALUES (3, 'admin112', 'pass', 1, 1, 0, 0, 0, 0);

[Jessica]

See, as I said like 85 times, you have some with a 0 and some with a 1 - you're looping through all of them, so you get each one.

Read the tutorial in the beginner section on mysql.

[forumnz]

Ooo i got it. Ok thanks. I have one other problem - how can i get it to display only one users info based on their unique id?

[Jessica]

use WHERE id=$id

Like I said, check out the tutorial.