javascript - Displaying images using for loop on PHP
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I am trying to display my images in a for loop in the form tag using PHP but it does not seem to work. I have also tried doing a return statement but nothing still appears.
<?php
$data = array("images/architecture-57e8d34a48_640.jpg",
"images/gateway-arch-57e2d64548_640.jpg",
"images/horseshoe-bend-57e6d6434f_640.jpg",
"images/lake-irene-57e6d24a4d_640.jpg",
"images/silhouette-57e8d5444e_640.jpg");
function displayCheckboxes(){
for ($i=0; $i<count($data); $i++) {
echo "<img src='".$data[$i] . "<br>";
}
}
?>
<body>
<main id ="main">
<form id="pics" action="process.php" method="get">
<label>Name: </label>
<?php echo displayCheckboxes();?>
</form>
</main>
</body>
Answer
Solution:
Maybe Like This:
<?php
$data = array("images/architecture-57e8d34a48_640.jpg",
"images/gateway-arch-57e2d64548_640.jpg",
"images/horseshoe-bend-57e6d6434f_640.jpg",
"images/lake-irene-57e6d24a4d_640.jpg",
"images/silhouette-57e8d5444e_640.jpg");
function displayCheckboxes(){
global $data;
for ($i=0; $i<count($data); $i++) {
echo "<img src='".$data[$i]."'>" . "<br>";
}
}
displayCheckboxes();
?>
Answer
Solution:
function displayCheckboxes($img_array){
foreach($img_array as $img) {
echo "<img src='".$img."'>'" . "<br>";
}
}
and
echo displayCheckboxes($data);
There is several errors with your function : - $data is not inside the function - .jpg is already in the image name.
Answer
Solution:
this should work,call the function with $data
$data = array("images/architecture-57e8d34a48_640.jpg",
"images/gateway-arch-57e2d64548_640.jpg",
"images/horseshoe-bend-57e6d6434f_640.jpg",
"images/lake-irene-57e6d24a4d_640.jpg",
"images/silhouette-57e8d5444e_640.jpg");
/* Write your displayCheckboxes() function here */
displayCheckboxes($data);
function displayCheckboxes($data){
for ($i=0; $i<count($data); $i++) {
echo "<img src='".$data[$i]."'.jpg>'" . "<br>";
}
}
?>
Answer
Solution:
<?php
$data = array(
"images/architecture-57e8d34a48_640.jpg",
"images/gateway-arch-57e2d64548_640.jpg",
"images/horseshoe-bend-57e6d6434f_640.jpg",
"images/lake-irene-57e6d24a4d_640.jpg",
"images/silhouette-57e8d5444e_640.jpg"
);
?>
<body>
<main id ="main">
<form id="pics" action="process.php" method="get">
<label>Name: </label>
<?php foreach($data as $image): ?>
<img src="<?= $image; ?>"></br>
<?php endforeach; ?>
</form>
</main>
</body>
This is an alternative way. Looks much cleaner and readable. And I actually don't think we need a function for printing HTML.
I would say NEVER put HTML inside PHP echo unless it is compulsory. The reason I am stating this is if you put HTML inside PHP, the code becomes easily messy. It becomes hard to understand the logic unless you are the one who coded. Particularly, if you are working with some designers or in future, they will have a hard time making even little changes.
Even I think using<?= than<?php echo is better.
Answer
Solution:
You are trying to give an extension ".jpg" of the images that you previously gave to it in the path of images in the array "$data".
For exemple:
We will take the first element in the array $data with index = 0 "$data[0]" (images/architecture-57e8d34a48_640.jpg):
In PHP looks like ->echo "<img src='".$data[0]."jpg'>" . "<br>";
,but in HTML looks like -><img src='images/architecture-57e8d34a48_640.jpg.jpg'>
You shold have to delete the extension ".jpg" from your echo to seem like:
echo "<img src='".$data[$i]."'>" . "<br>";
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