根据PHP条件触发JavaScript函数 [英] Trigger javascript function depending on PHP condition

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问题描述

我有一个表单将返回其URL中的值,包括表单是否有效
/contact2.php?result=contact-error-captcha&name=John
/ contact2。 php?result = contact-ok& name = John



我试图触发一个Google Analytics网页浏览量,具体取决于这个值,它通常在javascript中工作,如下所示:
onclick =ga('send','pageview','/ mypage');

这是我最后一次尝试,但似乎没有执行。

 < script type =text / javascript> 
函数error-captcha(){
alert(Hello!This works);
ga('send','pageview','/ pages / contact-error-captcha');

function contact-ok(){
ga('send','pageview','/ pages / contact-ok');
}
<?php if($ _GET ['result'] =='contact-error-captcha'):echoerror-captcha();;万一; ?>
<?php if($ _GET ['result'] =='contact-ok'):echocontact-ok();;万一; ?>
< / script>

我猜的问题是:
- PHP应该在JS之前(但是怎么做?)
- ga函数应该是on的东西(但是怎么样?)

我不是专业编码器,所以任何帮助表示赞赏。

解决方案

 < script type =text / javascript> 
函数error_captcha(){
alert(Hello!This works);
ga('send','pageview','/ pages / contact-error-captcha');
}
function contact_ok(){
ga('send','pageview','/ pages / contact-ok');
}
<?php if($ _GET ['result'] =='contact-error-captcha'):echoerror_captcha();;万一; ?>
<?php if($ _GET ['result'] =='contact-ok'):echocontact_ok();;万一; ?>
< / script>


I have a form that will return values in its URL, including whether the form is valid or not /contact2.php?result=contact-error-captcha&name=John /contact2.php?result=contact-ok&name=John

I am trying to trigger a Google Analytics Pageview depending on this value, which normally works in javascript like this: onclick="ga('send', 'pageview', '/mypage');"

Here is my last attempt, but it seems to not execute.

  <script type="text/javascript">
     function error-captcha() {
         alert("Hello! This works");
         ga('send', 'pageview', '/pages/contact-error-captcha');
     } 
     function contact-ok() {
         ga('send', 'pageview', '/pages/contact-ok');
     } 
     <?php if( $_GET['result'] == 'contact-error-captcha') : echo "error-captcha();"; endif; ?>
     <?php if( $_GET['result'] == 'contact-ok') :  echo "contact-ok();"; endif; ?>
 </script>

The problems I guess: - PHP should be before JS (but how?) - ga function should be "on" something (but how?)

I am not a pro coder, so any help appreciated.

解决方案

Issue in your function name not use '-' try this:

<script type="text/javascript">
     function error_captcha() {
         alert("Hello! This works");
         ga('send', 'pageview', '/pages/contact-error-captcha');
     } 
     function contact_ok() {
         ga('send', 'pageview', '/pages/contact-ok');
     } 
     <?php if( $_GET['result'] == 'contact-error-captcha') : echo "error_captcha();"; endif; ?>
     <?php if( $_GET['result'] == 'contact-ok') :  echo "contact_ok();"; endif; ?>
 </script>

这篇关于根据PHP条件触发JavaScript函数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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