首页 > 解决方案 > 无法将数据库中的字符串解码为 JSON

问题描述

我在stringify类似的帮助下将值存储在数据库中

var value = JSON.stringify(credit);

并通过ajax将其存储在数据库中。

javascript

$(".revolving_credit_row").each(function(){
        var revolving_credit_object = {};
        var that  = $(this);

        that.find('.revolving_credit_input').each(function (){
            var field_name = $(this).attr('data-field-name');
            var field_value = $(this).val();

            revolving_credit_object[field_name] = field_value;
        });

        revolving_credit.push(revolving_credit_object);
    });

var value = JSON.stringify(revolving_credit);

数据库中的值就像

[{"Creditor":"12500","Balance":"","Limit":"","Utilization":""},{"Creditor":"20525","Balance":"","Limit":"","Utilization":""},{"Creditor":"","Balance":"","Limit":"","Utilization":""}]

现在,当我开始检索时,它给了我确切的字符串,但我无法对其进行解码。

我正在使用这个功能。

$credit = $business[0]->credit;

$temp = json_decode($credit);

        print_r($temp);
        exit;

因为我print_r它没有给出任何输出,如果我使用var_export,或者var_dump它给我null。

标签: phpjson

解决方案


一定是:

$temp = '[{"Creditor":"12500","Balance":"","Limit":"","Utilization":""},{"Creditor":"20525","Balance":"","Limit":"","Utilization":""},{"Creditor":"","Balance":"","Limit":"","Utilization":""}]';

print_r($temp);

var_dump(json_decode($temp));
var_dump(json_decode($temp, true));

我在''之间没有这个,将无法使用该功能json_decode()


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