dtn43447 2013-04-14 02:19
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如何使用超链接“发布”样式调用php函数

I am currently trying to make a hyperlink that calls the same page that I am on, but a different PHP function. This seems like a simple enough solution however I don't want any information displayed in the URL. A "post" seems to be the best solution but I cannot find any results as to how to make this work.

<?php 
function myFirst(){ 
    echo 'The First ran successfully.'; 
}
function mySecond(){ 
    echo 'The Second ran successfully.'; 
}
?> 

<html><body> 

<?php 

if (isset($_GET['run'])){
    $linkchoice=$_GET['run'];
}else{
    $linkchoice='';
}

switch($linkchoice){ 
    case 'first' : 
        myFirst(); 
        break; 

    case 'second' : 
        mySecond(); 
        break; 

    default : 
        echo 'no run'; 
} 

?> 

<hr> 
<a href="?run=first">Link to First</a> 
<br/> 
<a href="?run=second">Link to Second</a> 
<br/> 
<a href="?run=0">Refresh No run</a> 

</body></html>
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2条回答 默认 最新

  • dongsuyou6938 2013-04-14 02:29
    关注

    If u want to use POST by pressing something in the page, u can either use JS to turn that into a POST request, or simply submit a form.

    Assuming u use jQuery

    <a href="baba/was/here/this/is/the/url" onclick="manda_mi()">go somewhere</a>
    <form id="koko" style="display:none" target="baba/was/here/this/is/the/url" method="post">
    <input type="hidden" name="run" value="boo" />
    </form>
    ...
    ...
    function manda_mi(){
      $('#koko').submit();
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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