[英]How can i add two variables between single quotes
I have我有
function delImage(posteId,imagebinId) {
$.get("<?php echo base_url('/Home/deletePostimage/') ?>");
return false;
}
i want to be like /Home/deletePostimage/posteId/imagebinId我想像 /Home/deletePostimage/posteId/imagebinId
i tried我试过
$.get("<?php echo base_url('/Home/deletePostimage/')+posteId+"/"+imagebinId ?>");
but it divide the two numbers但它将两个数字相除
like posteId=5 imagebinId =2像 posteId=5 imagebinId =2
the results will be结果将是
/Home/deletePostimage/2,5
/首页/删除帖子图片/2,5
the params from来自的参数
<a class="postimgsd" onClick="if(confirm('Are you sure you want to delete this image ?')){delImage(<?php echo $value['id'],$get_images[0]['binId']; ?>); rmItem(<?php echo $i; ?>);}else{} ; " >
With ES6, you can do this with string interpolation :在 ES6 中,你可以通过字符串插值来做到这一点:
$.get(`<?php echo base_url('/Home/deletePostimage/${postId}/${imagebinId}') ?>`);
Without ES6, there is another answer that answers it, or use this:如果没有 ES6,还有另一个答案可以回答它,或者使用这个:
var url = '/Home/deletePostimage/' + postId + '/' + imagebinId;
$.get("<?php echo base_url('" + url + "') ?>");
Furthermore, the code that calls this function should actually be:此外,调用此函数的代码实际上应该是:
<a class="postimgsd" onClick="checkDeletion()" />
<script type='text/javascript'>
function checkDeletion() {
if(confirm('Are you sure you want to delete this image ?')) {
var postId = <?php echo $value['id']; ?>;
var imagebinId = <?php echo $get_images[0]['binId']; ?>;
delImage(postId, imagebinId);
rmItem(<?php echo $i; ?>);
}
}
</script>
You need to keep in mind that PHP is processed on back-end before send the html to your user browser.您需要记住,在将 html 发送到您的用户浏览器之前,PHP 是在后端处理的。 That said, everything between php tags (
<?php ... ?>
) is rendered/processed before javascript even exists.也就是说,php 标记(
<?php ... ?>
)之间的所有内容都在 javascript 存在之前呈现/处理。
So if you put any JavaScript code inside your PHP code it won't work.因此,如果您在 PHP 代码中放入任何 JavaScript 代码,它将无法工作。
To make it work and be clean, do something like this:要使其正常工作并保持清洁,请执行以下操作:
function delImage(posteId,imagebinId) {
let url = "<?php echo base_url('/Home/deletePostimage/') ?>";
$.get(url + posteId + "," + imagebinId);
return false;
}
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