javascript - Resource interpreted as Document but tran..

The question is as mentioned. When I tested the data, the browser popped up the prompt Resource interpreted as Document but transferred with MIME type application/json: "http://localhost:82/js4-1.php?type=findnew&data=pic". ,is the json data type wrong?

php code:
error_reporting(0);
header("Content-type:application/json;charset=utf-8;");
//Test data connection
$icon=mysql_connect("localhost ","root","");
if(!$icon){

<code>//失败
die('Could not connect:'.mysql_error());</code>

}
//Connect to the database
mysql_select_db("news_data",$icon);
//Set Chinese
mysql_query("set names utf8");
//Get the front-end type, data parameters
$type=@$_GET[ 'type'];
$sleword=@$_GET['data'];
switch ($type) {

<code>//从主页前段到php
case 'findnew':
    $sql="select id,new_title,new_class from info_look";//创建查询语句
    $re=mysql_query($sql);//执行语句
    $wordArray=[];//创建空字符串承载获取到的数据
    while($row=mysql_fetch_array($re)){
        $wordArray[]=$row;//将结果赋值给数组
    }
    //下面打算用循环找到对应传递过来的data参数，匹配到了，就把他回传给前台
     for($i=0;$i<count($wordArray);$i++){//php数组没有length,获取数组长度要用count函数?
             if($wordArray[i].new_class==$sleword){
                 //这里传值$wordArray[i].new_class的值给js
                 //...    
                 echo json_encode ($wordArray[i]['new_class']);
             }
         }
    break;
//从后台前端网页插入数据到php
case insertdata:
    $sql="INSERT INTO info_look (new_title,new_class) values ('$_POST[new_title_info]','$_POST[new_class_info]')";
    if(!mysql_query($sql,$icon)){
        die('Error:' . mysql_error());
    }
    echo "one title added!";
    break;</code>

}
mysql_close($icon);
?>

Reply content:

The question is as mentioned. When I tested the data, the browser popped up the prompt Resource interpreted as Document but transferred with MIME type application/json: "http://localhost:82/js4-1.php?type=findnew&data=pic". ,is the json data type wrong?

php code:
error_reporting(0);
header("Content-type:application/json;charset=utf-8;");
//Test data connection
$icon=mysql_connect("localhost ","root","");
if(!$icon){

<code>//失败
die('Could not connect:'.mysql_error());</code>

}
//Connect to the database
mysql_select_db("news_data",$icon);
//Set Chinese
mysql_query("set names utf8");
//Get the front-end type, data parameters
$type=@$_GET[ 'type'];
$sleword=@$_GET['data'];
switch ($type) {

<code>//从主页前段到php
case 'findnew':
    $sql="select id,new_title,new_class from info_look";//创建查询语句
    $re=mysql_query($sql);//执行语句
    $wordArray=[];//创建空字符串承载获取到的数据
    while($row=mysql_fetch_array($re)){
        $wordArray[]=$row;//将结果赋值给数组
    }
    //下面打算用循环找到对应传递过来的data参数，匹配到了，就把他回传给前台
     for($i=0;$i<count($wordArray);$i++){//php数组没有length,获取数组长度要用count函数?
             if($wordArray[i].new_class==$sleword){
                 //这里传值$wordArray[i].new_class的值给js
                 //...    
                 echo json_encode ($wordArray[i]['new_class']);
             }
         }
    break;
//从后台前端网页插入数据到php
case insertdata:
    $sql="INSERT INTO info_look (new_title,new_class) values ('$_POST[new_title_info]','$_POST[new_class_info]')";
    if(!mysql_query($sql,$icon)){
        die('Error:' . mysql_error());
    }
    echo "one title added!";
    break;</code>

}
mysql_close($icon);
?>

The problem lies in the last sentence. The previous paragraph expects to return 'json'. Although you set the header in the pho header, you can see this sentence

''Php

<code>echo "one title added!";</code>

'''