如何使用POST方法从数组中获取值 [英] How to get values from array using POST method
本文介绍了如何使用POST方法从数组中获取值的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
如何使用POST方法在另一个页面中获取值?
< label>
< input type =radiovalue =1name =jsq [1]checked> 1
< / label>
< label>
< input type =radiovalue =2name =jsq [1]> 2
< / label>< / br>
< label>
< input type =radiovalue =3name =jsq [1]> 3
< / label>
< label>
< input type =radiovalue =4name =jsq [1]> 4
< / label>< / br>
< label>
< input type =radiovalue =5name =jsq [1]> 5
< / label>
我想获得 jsq [1]
在 myarr [1]
中。我怎么能这样做?
$ myarr [1] = $ _POST ['jsq [1]'];
解决方案
jsq [1]不只是jsq ...
你几乎在那里,但应该是
$ myarr [1] = $ _POST ['jsq'] [1];
使用print_r($ _ POST)检查这些事情的方法总是一个好主意;或var_dump($ _ POST);
How to get value in another page using the POST method?
<label>
<input type="radio" value = "1" name="jsq[1]" checked>1
</label>
<label>
<input type="radio" value = "2" name="jsq[1]">2
</label></br>
<label>
<input type="radio" value = "3" name="jsq[1]">3
</label>
<label>
<input type="radio" value = "4" name="jsq[1]">4
</label></br>
<label>
<input type="radio" value = "5" name="jsq[1]">5
</label>
I want to get value of jsq[1]
in myarr[1]
. How can I do that?
$myarr[1] = $_POST['jsq[1]'];
解决方案
Without going into the question why you're naming them jsq[1] an not just jsq...
you're almost there, but it should be
$myarr[1] = $_POST['jsq'][1];
It's by the way always a good idea to check this king of things using print_r($_POST); or var_dump($_POST);
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