php get 改 post的疑问
然后在picshow_menu.php中用 $picidxnumber=$_GET["picnum"]; 来拿到值.
现在想把这个get改成post的方法,picshow_menu.php中我知道是要改成$picidxnumber=$_POST["picnum"]; 但是不知道src这里要怎么改.我要怎么把这个picnum用post的方法发送出去呢?
请大家指点一下,谢谢了.
------解决方案--------------------
把提交的FORM表单中的method="get" 改成 method="post"
------解决方案--------------------
要么已表单的形式的提交啊。。。或者 jquery. $.post("picshow_menu",{picnum:XXXX},function(d){});
这样提交。。。但是我觉得 我回答的东西 对不上你想做的事...
------解决方案--------------------
那是form的方法。或者ajax时,使用POST形式。
------解决方案--------------------
你如果是不想让别人看到你的url传值可以加密url的传值,这样就没必要去改提交方式了
另外,如果你真想要改的话,上面的同道都已经给出了正确的答案,ajax和表单提交两种方法都可以的,这就看你自己的选择了

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