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@NogDogNov 18.2016 — #Does getCar_1() and getCar_2() actually output HTML? If so, then trying to assign whatever it actually returns (if anything?) to those $getCar_[1|2] variables might be messing up your HTML by outputting whatever it does right at that point in the code. @NogDogNov 19.2016 — #Yeah, that's the problem with functions actually echoing output: you can't call them to test their result or modify the output. If you want to test the result then use it later, you would modify that function to return the text instead of outputting it. You could do something like:
Then you can call the function and assign it's output to a variable like you did, earlier, or if you just want to use it directly, echo the function.
CSS not applying to div when adding PHP condition
Hello!
In the first place, i’m sorry if my question mathes the previous one, they are quite similar. I try to adapt the answer from the previous question for my problem but it didn’t work.
So here is the thing.
I want to conditionally display a div so when there are no values, the div is not displayed. Without the php if part, all is working fine, the css styles are applying to the divs. But when i add the if condition, the CSS doesn’t apply anymore.
Here is the div part:
[CODE]<?php
$getCar_1 = getCar_1();
$getCar_2 = getCar_2();
if(!empty($getCar_1) || !empty($getCar_2)){
?>
<div class=”block”>
<div class=”up”>
<div class=”title”><h4>Cars</h4></div>
</div>
<div class=”down”>
<div class=”column_1″><b><?php getCar_1(); ?></b></div>
<div class=”column_2″><b><?php getCar_2(); ?></b></div>
</div>
</div>
<?php
}
?>
and here is the CSS:
[CODE].block{
width:900px;
margin:0 auto;
overflow:hidden;
float:left;
}
.up{
width:100%;
text-align:center;
border:1px solid;
margin-top:10px;
margin-bottom:10px;
background-color:skyblue;
}
.down{
width:100%;
clear:both;
}
.column_1{
float:left;
width:47%;
padding:10px;
}
.column_2{
float:right;
width:47%;
text-align:right;
padding:10px;
}
If i use only this code:
[CODE]<div class=”block”>
<div class=”up”>
<div class=”title”><h4>Cars</h4></div>
</div>
<div class=”down”>
<div class=”column_1″><b><?php getCar_1(); ?></b></div>
<div class=”column_2″><b><?php getCar_2(); ?></b></div>
</div>
</div>
it displays normal. Here is a picture:
[URL=”https://i.stack.imgur.com/fj1RH.jpg”]https://i.stack.imgur.com/fj1RH.jpg
And when adding the php if part it displays this:
[URL=”https://i.stack.imgur.com/dMPUY.jpg”]https://i.stack.imgur.com/dMPUY.jpg
So when i add the php if, the css has no effect anymore. I tried many variants but did not find the solution.
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@NogDogNov 18.2016 — #0
@NogDogNov 19.2016 — #Yeah, that's the problem with functions actually echoing output: you can't call them to test their result or modify the output. If you want to test the result then use it later, you would modify that function to return the text instead of outputting it. You could do something like:[code=php]
<?php
function getCar_2(){
global $connect;
$get_car_2 = "SELECT * FROM cars WHERE category_id=5";
$get_car_result_2 = mysqli_query($connect, $get_car_2) or die(mysqli_query());
$text = '';
while($row_car_2 = mysqli_fetch_array($get_car_result_2)){
$car_model_2 = $row_car_2['car_name'];
$car_image_2 = $row_car_2['car_image'];
if(isset($car_model_2) && isset($car_image_2)){
$text = "
<div id='' style='white-space:nowrap;'>
<p id='model' style='display:inline-block; margin-right:10px; margin-left:10px; padding-top:10px; position:relative; bottom:35px;'>$car_model_2</p></a>
<img src='administrator/images/$car_image_2' width='120' height='80' style='display: inline-block; box-shadow: 0 0 11px #000;'/><br>
</div>";
}
}
return $text;
}
?>
[/code]Then you can call the function and assign it's output to a variable like you did, earlier, or if you just want to use it directly, echo the function.