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我已创建一个自定义警告对话框,但我不能将它调用PHP无法调用自定义警报的JavaScript函数在PHP
下面是警告对话框的代码
<script>
function CustomAlert(){
this.render = function(dialog){
var winW = window.innerWidth;
var winH = window.innerHeight;
var dialogoverlay = document.getElementById('dialogoverlay');
var dialogbox = document.getElementById('dialogbox');
dialogoverlay.style.display = "block";
dialogoverlay.style.height = winH+"px";
dialogbox.style.left = (winW/2) - (550 * .5)+"px";//position of dialog box
dialogbox.style.top = "200px"; //height of the box
dialogbox.style.display = "block";
document.getElementById('dialogboxhead').innerHTML = "Registraion Succesfull";
document.getElementById('dialogboxbody').innerHTML = dialog;
document.getElementById('dialogboxfoot').innerHTML = '<button onclick="Alert.ok()">OK</button>';
}
this.ok = function(){
window.open("employer/personal","_self");
}
}var Alert = new CustomAlert();
下面
是PHP代码
<?php
echo "<script> CustomAlert(Alert.render('You look very pretty today.')); </script>";
?>
您无法在PHP中调用JavaScript函数。在JavaScript运行时,PHP已经完成。为了运行它,你需要捕获一个用户事件([见addEventListener](http://stackoverflow.com/questions/6348494/addeventlistener-vs-onclick)),例如点击按钮,滚动到特定位置,点击输入字段等。 – gibberish
您的PHP文件是否还回显或包含定义您的CustomAlert和Alert对象的脚本块,还是只回显单行?换句话说,如果在渲染后查看页面的源代码,它看起来像什么? – stratedge