如何选择具有相同名称的所有输入并按名称对其进行索引 [英] How to select all inputs having same name and index them by their name
问题描述
我必须处理由第三部分生成的表单。此表单提供了一些复选框多选属性。但输入名称是相同的,并且PHP $ _POST无法处理相同的键(仅处理最后一个数据)。
I have to handle a form generated by a third part. This form provides some checkbox "multi-select" attributes. But input name's are identical, and PHP $_POST can't handle identical keys (only the last data is handled).
以下是表单的示例:
<form>
<input type="checkbox" name="attr1" value="1" />
<input type="checkbox" name="attr1" value="2" />
<input type="checkbox" name="attr1" value="3" />
<input type="checkbox" name="attr1" value="4" />
<input type="checkbox" name="attr2" value="xx" />
<input type="checkbox" name="attr3" value="a" />
<input type="checkbox" name="attr3" value="b" />
<input type="checkbox" name="attr3" value="c" />
<input type="checkbox" name="attr3" value="d" />
<input type="text" name="attr4" value="" />
</form>
我想解析所有复选框输入,只获取同名输入组,并追加 []以他们的名字......
I'd like to parse all checkbox inputs, get only group of same name inputs, and append "[]" to their name...
我想在jQuery处理后获得:
What I want to get after jQuery processing :
<form>
<input type="checkbox" name="attr1[]" value="1" />
<input type="checkbox" name="attr1[]" value="2" />
<input type="checkbox" name="attr1[]" value="3" />
<input type="checkbox" name="attr1[]" value="4" />
<input type="checkbox" name="attr2" value="xx" />
<input type="checkbox" name="attr3[]" value="a" />
<input type="checkbox" name="attr3[]" value="b" />
<input type="checkbox" name="attr3[]" value="c" />
<input type="checkbox" name="attr3[]" value="d" />
<input type="text" name="attr4" value="" />
</form>
有没有办法识别同名的输入组?
Is there a way to identify group of inputs with same name ?
推荐答案
如果存在另一个具有相同项目的项目,则会将 []
附加到每个复选框项的名称name。
This will append []
to the name of each checkbox item if there exists another item with the same name.
var name_map = {};
$("input[type=checkbox]") // for all checkboxes
.each(function() { // first pass, create name mapping
var name = this.name;
name_map[name] = (name_map[name]) ? name + "[]" : name;
})
.each(function() { // replace name based on mapping
this.name = name_map[this.name];
});
演示: http://jsfiddle.net/gnfsG/
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