Question

PHP首先在JS之前执行，但如果数据不是空的，就像在这个条件PHP语句中一样，有一种简单的方法可以输出确认对话框（如Javascript确认）：

<?php
if (!empty($data)) {

//data is not empty
//Ideally I want to show a confirmation dialog in this part to the user
//to confirm if the user wants to overwrite the existing data

//If the user confirms
//Overwrite the existing data here

} else {

//data is empty,proceed to other task
//don't show any dialog

}

我不想向用户添加另一个表单，只是像JS确认的简单确认对话框。或者是否有PHP替换函数？感谢。

Answer 1

我喜欢将数据和代码分开，所以我喜欢从PHP

创建一个JS变量

// I always use json_encode when creating JS variables from PHP to avoid
// encoding issues with new lines and quotes.
var isDataEmpty = <?php echo json_encode(empty($data)); ?>;

if (!isDataEmpty) {
   if (confirm("Some message")) {
      overwriteExistingData();
   }
} else {
   // proceed
}

Answer 2

您可以让您的PHP代码编写如下代码的JavaScript：

<script type="JavaScript">
$(document).ready(function(){
    <?php
    if (!empty($data)) {
    ?>
    var answer = confirm("Are you sure you want to overwrite?");
    <?php
    } else {
    ?>
    // Data is empty
    <?php
    }
    ?>
});
</script>

Answer 3

请参阅以下代码

<?php
if (!empty($data)) {
    echo '<script language="javascript" type="text/javascript">window.confirm("data is not empty")</script>';

//data is not empty
//Ideally I want to show a confirmation dialog in this part to the user
//to confirm if the user wants to overwrite the existing data

//If the user confirms
//Overwrite the existing data here

} else {
    echo '<script language="javascript" type="text/javascript">window.confirm("data is empty here")</script>';

//data is empty,proceed to other task
//don't show any dialog

}

如何在条件PHP语句中使用JS确认？

3 个答案: