Question

我在将多维数组[N] [N]从javascript发送到PHP时遇到了一些麻烦，我已经尝试了很多我在这里找到的解决方案，但我不知道我做错了什么。

我的JQuery代码（从HTML表中保存值）：

$rowArray = {};

for ($i = 0; $i < $myRows.length; $i++) {
    $row = $($myRows[$i]).find('td');
    $rowArray[$i] = {};
    for ($j = 0; $j < $row.length - 1; $j++) {
        $rowArray[$i][$j] = $($row[$j]).html();
    }
}

然后：

$myJsonString = JSON.stringify($rowArray);

$.ajax({
    type: "POST",
    url:"../../download/myStore.php",
    data: { table: $myJsonString },
    success: function(data){
        console.log(data);
    }
});

PHP方面：

echo $_POST['table']; //just to see what is coming, but i want to work as array

//$data = json_decode($_POST['table'],true); -> when I echo $data, the output is an error Array to String conversion

输出：

{"0":{"0":"Cadastrado em","1":"Data da Venda","2":"Empreendimento","3":"Bloco/<br>Unidade","4":"Cliente/<br>Parceiro","5":"Valor","6":"Filial","7":"Gerente","8":"Corretor","9":"Veículo"},"1":{"0":"27/04/2016","1":"11/04/2016","2":"Villa Flora Hortolândia - Cond. 06","3":"Bloco/Torre: 13, Unidade: 283","4":"Lidiane Sasaki Santana","5":"20.664.259","6":"Campinas","7":"NATAL","8":"WILLIAM PILOTO","9":"Internet"},"2":{"0":"12/04/2016","1":"12/04/2016","2":"Lifespace Curitiba","3":"Bloco/Torre: 1, Unidade: 2404","4":"ANA","5":"351.000","6":"Curitiba","7":"André Barbosa de Lima","8":"Daniele","9":"Google"},"3":{"0":"12/04/2016","1":"12/04/2016","2":"ROSSI ATUAL MORADA","3":"Bloco/Torre: 3, Unidade: 407","4":"BERNADETE STARKE","5":"245.000","6":"Curitiba","7":"André Barbosa de Lima","8":"Dranka","9":"Google"},"4":{"0":"12/04/2016","1":"12/04/2016","2":"Lifespace Curitiba","3":"Bloco/Torre: 2, Unidade: 1105","4":"FLAVIA AMARAL","5":"272.500","6":"Curitiba","7":"André Barbosa de Lima","8":"Jesus","9":"Yahoo"},"5":{"0":"12/04/2016","1":"12/04/2016","2":"Lifespace Curitiba","3":"Bloco/Torre: 2, Unidade: 1809","4":"itajana","5":"270.500","6":"Curitiba","7":"André Barbosa de Lima","8":"Daniele","9":"Site Rossi"},"6":{"0":"27/04/2016","1":"14/04/2016","2":"Villa Flora Hortolândia - Cond. 05","3":"Bloco/Torre: 5, Unidade: 41","4":"Andre Fernando Da Silva Gradino","5":"184.303","6":"Campinas","7":"NATAL","8":"TIAGO","9":"Cadastro Manual"}}

如何访问每个数组中的每个索引以获取我的值？
每当我尝试使用像$_POST['table'][0]这样的索引时，输出就是'{' 对不起我的英语并对这个noob问题感到抱歉，但我已经在这里停留了好几个小时，并且已经在谷歌和谷歌上尝试了很多解决方案。

Answer 1

正如你已经在你的代码里面（但已被注释掉）你可以使用json_decode($_POST['table'],true);当你使用这个函数时，你得到一个数组，有7个（在这个例子中）。

像这样：

$data = json_decode($_POST['table'], true);

发生错误，因为您echo变量。使用var_dump($data)来查看它真的是一个数组。 $data[0]，$data[1]，...然后保存数据。

编辑：因为你有嵌套数组，你必须再次访问子数组，如下所示：

$data = json_decode($_POST['table'], true);
$var = $data[0][0]; // holds "Cadastrado em"
$var2 = $data[0][1]; // holds "Data da Venda"

我不确定，如果是这样的话，你在数组中有一个数组。

Answer 2

使用json_decode将字符串文字转换为PHP对象。

 $json = '["apple","orange","banana","strawberry"]';
 $ar = json_decode($json);
 // access first element of $ar array
 echo $ar[0]; // apple

Answer 3

您的注释行

department_id

很好，但是你不能回显数组。这将失败：

   def allocation_substitute(self,cr,uid,ids,roster_day,context=None):
        sub_day=vals.get(roster_day)
        sub_time_slot=vals.get(time_slot)

        allocation_obj=self.pool.get('roster.allocation')
        original_employee_id =  allocation_obj.browse(cr,uid, values['emp_id']).id
        original_employee_roster=allocation_obj.browse(cr,uid, original_employee_id).roster_type

        emp_id = 1  # Search from hr.employee 
        department_id = 1  # Search from hr.department

        values={'emp_id': emp_id,
                'department_id': department_id,
                }
        allocation_id=allocation_obj.create(cr, uid, value, context=context)  

        return True

这将有效：

$data = json_decode($_POST['table'],true);

从Jquery发送多维数组到PHP。我做错了什么？

3 个答案: