Php iam返回json_encode格式的数据，并尝试用ajax记录输出，但出现解析错误。？

我的模型--我从Mysql数据库获取数据并将其返回给控制器

public function getCharges(){
    $where = "charge_type  = 'variable';
    $data = $this->read('charges',$where);
    return $data;
}

---

我的控制器——这里我将json格式的数据编码为ajax

if (isset($_POST['method']) &&  $_POST['method'] == 'getCharges'){
     $this->load->model('Apiweb_m');
     $data = $this->Apiweb_m->getCharges();
     echo json_encode($data);
}

---

我的阿贾克斯

var method = "getCharges";
$.ajax({
    type: 'POST',
    data: 'method='+method,
    url : api_url+"apiweb",
   /* dataType: "json",*/
    success:function(msg){
        var data =  $.parseJSON(msg);
        console.log(data);// here iam getting Parse error 
    },
    error:function(Xhr, status, error){
        console.log(Xhr);
        console.log(status);
        console.log(error);
    }

---

将模型中的功能更改为：

public function getCharges(){
    $where = "charge_type  = 'variable'";
    $data = $this->read('charges',$where);
    return $data;
}

和ajax：

var method = "getCharges";
$.ajax({
    type: 'POST',
    data: {method:method},
    url : api_url+"apiweb",
    dataType: "json",
    success:function(msg){
        var data =  $.parseJSON(msg);
        console.log(data);// here iam getting Parse error 
    },
    error:function(Xhr, status, error){
        console.log(Xhr);
        console.log(status);
        console.log(error);
    }

---

您必须设置数据类型ajax属性json。希望您能这样做

var method = "getCharges";
$.ajax({
    type: 'POST',
    data: 'method='+method,
    url : api_url+"apiweb",
    dataType: "json",
    success:function(msg){
        var data =  $.parseJSON(msg);
        console.log(data);// here iam getting Parse error 
    },
    error:function(Xhr, status, error){
        console.log(Xhr);
        console.log(status);
        console.log(error);
    }

---

请添加您得到的示例ajax响应，另外，如果它的响应正确，则

contentType:'application/json'

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