用JQuery Ajax提交特定的表单 [英] Submit a specific form with JQuery Ajax
本文介绍了用JQuery Ajax提交特定的表单的处理方法,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
我在页面上使用了许多相同ID的表单。当我按任何形式的提交按钮时,首先第一个将提交,第二个单击第二个......之后。但是当我按下提交按钮时,表单会提交按钮所属的位置。
I had many forms with the same id on a page. When I press the submit button of any form, first the first will submit, after the second click the second ... . But I want when I press the submit button, the form will submit where the button belongs to. How can I do that.
在这里我的JS代码:
$(document).on('submit','#ajax_form',function(e) {
var form = $('#ajax_form');
var data = form.serialize();
$.post('game/write.php', data, function(response) {
console.log(response);
$('#power').replaceWith(response);
});
return false;
});
这里的HTMl代码:
<div id="power">
<div class="span4">
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="1" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="4" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="7" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
</div>
<div class="span4">
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="2" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="5" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="8" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
</div>
<div class="span4">
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="3" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="6" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
<form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="9" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form>
</div>
推荐答案
试试这个:
Try this:
$(document).on('click','button.btn',function(e) {
//you will trigger this function when you click a button
//this will select the parent, i.e., the form
var form = $(this).parent();
var data = form.serialize();
$.post('game/write.php', data, function(response) {
console.log(response);
$('#power').replaceWith(response);
});
return false;
});
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