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我serializeArray用来获取所有元素，我得到了像这样的对象

[{name: "code[1][barcode]", value: "45534"},
{name: "code[1][rf_id]", value: "535353"},
{name: "code[1][serialize]", value: ""},
{name: "code[2][barcode]", value: "45534"},
{name: "code[2][rf_id]", value: "535353"},
{name: "code[2][serialize]", value: ""},
{name: "custodian[]", value: "3"},
{name: "custodian[]", value: "4"},
{name: "custodian[]", value: "5"}]

我想像这样转换它

{
    code:[
            {barcode:"45534",rf_id:"535353",serialize:""},
            {barcode:"45534",rf_id:"535353",serialize:""}
        ],
    custodian: [3,4,5]
}

目前我正在使用这个脚本

var x = $('form#acquiredetail').serializeArray();
console.log(x);
var formData = {};
$.each(x, function(i, field){
    if(field.value.trim() != ""){
      formData[field.name] = field.value;
    }
});

并获得输出

虽然我能够code正确地获得价值，但在后端/Laravel 但问题在于custodian，我得到了最后一个价值，

custodian[]:"5"

我怎样才能解决这个问题。或者有什么更好的解决方案？

我的目标是使用简短而通用的代码将所有元素值传递给 php。请建议是否有任何替代方法。

4

2 回答 2

2

您必须更改序列化以查看元素是否以数组指示符结尾[]并根据元素的数量重建键：

var x = [{name: "code[1][barcode]", value: "45534"},
{name: "code[1][rf_id]", value: "535353"},
{name: "code[1][serialize]", value: ""},
{name: "code[2][barcode]", value: "45534"},
{name: "code[2][rf_id]", value: "535353"},
{name: "code[2][serialize]", value: ""},
{name: "custodian[]", value: "3"},
{name: "custodian[]", value: "4"},
{name: "custodian[]", value: "5"}];


var formData = {};
var formDataArrays = {};

$.each(x, function(i, field){
    if(field.value.trim() != ""){
      if (/\[\]$/.test(field.name)) {
        var fName = field.name.substr(0,field.name.length-2);
        if (!formDataArrays[fName]) {
          formDataArrays[fName] = [];
        }
        formData[fName+"["+formDataArrays[fName].length+"]"] = field.value;
        formDataArrays[fName].push(field.value);
      } else {
        formData[field.name] = field.value;
      }
    }
});

console.info(formData);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

于 2017-12-01T10:48:01.670 回答
2

运行代码片段并享受:) 

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
var x = 
[{name: "code[1][barcode]", value: "45534"},
{name: "code[1][rf_id]", value: "535353"},
{name: "code[1][serialize]", value: ""},
{name: "code[2][barcode]", value: "45534"},
{name: "code[2][rf_id]", value: "535353"},
{name: "code[2][serialize]", value: ""},
{name: "custodian[]", value: "3"},
{name: "custodian[]", value: "4"},
{name: "custodian[]", value: "5"}];

   var formData = {};
    $.each(x, function(i, field){
    	if(field.value.trim() != ""){
    		if(formData[field.name] != undefined){
    			var val = formData[field.name];
    			if(!Array.isArray(val)){
    				 arr = [val];
    			}
    			arr.push(field.value.trim());
    			formData[field.name] = arr;
    		}else{
    		  formData[field.name] = field.value;
    		}
        }
    });
    console.log(formData );
    </script>

于 2017-12-01T11:16:20.583 回答