本文介绍了将值从 PHP 传递到 JavaScript的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试获取 mysql 数据库中的 id 值.但是,每次单击图像时,都会得到 null.我使用 this 来获取值,但它不起作用,因为它一直在警告框中给我 null.
I'm trying to get the value which is the id in the mysql database. However, each time I click on the image, I get null. I used this to get the value but it is not working as it keeps giving me null in the alert box.
<!DOCTYPE html> <html> <head> <title></title> </head> <body> <?php mysql_connect('localhost','root',''); mysql_select_db("ajax"); $query="SELECT * FROM xxxx"; $result= mysql_query($query); while($row= mysql_fetch_array($result)){ echo "<img src='".$row['filepath']."' value='".$row['ID']."' id='".$row['ID']."' onclick='getrating(this.value);'>"; echo "<br>"; } ?> <script type="text/javascript" > function getrating(row_id){ var x = document.getElementById(row_id); alert(x); } </script> </body> </html>有什么问题?
推荐答案你需要 getrating(this.id) 代替.图像没有 value 属性.
You need getrating(this.id) instead. Images don't have a value property.
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将值从 PHP 传递到 JavaScript
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