如何将php变量作为参数传递给JavaScript函数
[英]How to pass php variable to JavaScript function as parameter
问题描述
I want to pass php image url as javascript function parameter. 
我想将php图片网址作为javascript函数参数传递。
Please let me know how can I do this. 请让我知道我该怎么做。
<html> <?php $image=$row['image']; $img=$loc.'user/'.$image; ?> <script type=text/javascript> function demo() { some code here; } </script> <body> <button type="submit" onclick=demo(<?php echo $img?>) > </body> </html>
In javascript I am doing some editing work on image . 在javascript中，我正在对图片进行一些编辑工作。
How can I pass this php variable (ie image url as javascript function parameter) 如何传递此php变量（即，图像url作为javascript函数参数）
4 个解决方案
解决方案1
2 2015-11-04 11:54:18
You need to enclose the PHP code within quotes like as 您需要将PHP代码用quotes例如
<button type="submit" onclick="demo('<?php echo $imageurl;?>');" >
//^^ ^^
Within your Javascript as you were passing value within function but not capturing that value within your function demo() 当您在函数中传递值但未在function demo()捕获该值时，在Javascript中
function demo(obj)
{ //^^^ Capturing value within function using variable name obj
alert(obj);
}
解决方案2
2 2015-11-04 11:57:38
您必须用引号将其包装，并且必须关闭button
<button type="submit" onclick='demo("<?php echo $imageurl?>")' ></button>
解决方案3
0 2015-11-04 11:56:23
<button type="submit" onclick="demo('<?php echo $img;?>');" >
解决方案4
0 2015-11-04 11:59:42
I found $img is your variable and in code you are trying to send $imageurl 我发现$img是您的变量，并且在代码中您尝试发送$imageurl
<html>
<?php
$image=$row['image'];
$img=$loc.'user/'.$image;
?>
<script>
function demo(val1)
{
alert(val1);
}
</script>
<body>
<button type="submit" onclick=demo('<?php echo $img?>') >Click Me</button>
</body>
</html>
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