changing the style inside "if" statement
I was trying to change the style of only a part of php. This is my codes;
if($fetch_array)
{
$foto_destination = $fetch_array['foto'];
echo "<img src = '$foto_destination' height='150px' width='150px'>";
}
else
{
?&g开发者_高级运维t;
<div style= "position:absolute; left:350px; top:70px;">
<?php
echo "<img src = 'images/avatar_default.png' height='150px' width='150px'>";
?>
</div>
But, this php part is inside if. This is why i could not change it? I want to display the image where i define it inside the div tag if the statement of "if" is true. How can i do this? Where am i missing? Thanks
If I understand you correctly, it should be:
<?php
if($fetch_array){
?>
<div style= "position:absolute; left:350px; top:70px;">
<?php
$foto_destination = $fetch_array['foto'];
print " <img src = '$foto_destination' height='150px' width='150px'>";
}else{
?>
<div style= "position:absolute; left:350px; top:70px;">
<img src = 'images/avatar_default.png' height='150px' width='150px'>
<?php
}
?>
</div>
It shows the $foto_destination, if there is one.
HTH
Did you mean like this?
<?php
if($fetch_array) {
$photo = $fetch_array['foto'];
$styles = 'position:absolute; left:350px; top:70px;';
} else {
$photo = 'images/avatar_default.png';
$styles = 'position:absolute; left:350px; top:70px;';
}
?>
<div style="<?php echo $styles; ?>">
<img src="<?php echo $photo; ?>" height="150" width="150" />
</div>
That is correct or you can do an isset()
if (isset($fetch_array) {
...
The only advantage being that it will not error if the variable is undefined
Here's a more compact version, shorttags must be enabled.
<div style="position:absolute; left:350px; top:70px;">
<img src="<?= isset($fetch_array['foto']) ? "images/avatar_default.png" : $foto_destination['foto'] ?>" height="150px" width="150px" />
</div>
otherwise:
<div style="position:absolute; left:350px; top:70px;">
<img src="<?php echo isset($fetch_array['foto']) ? "images/avatar_default.png" : $foto_destination['foto'] ?>" height="150px" width="150px" />
</div>
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