如何从运行PHP和MySQL的Web服务器读取JSON数据？ [英] How to reads JSON data from a web server running PHP and MySQL?

问题描述

我尝试从运行PHP和MySQL的Web服务器读取JSON数据。

我从 http://www.w3schools.com/json/json_example.asp

然后我用我的urlm改变了他们的网址http://localhost/demo.php

但它不起作用。我是JSON的初学者，我不知道我的代码发生了什么。请帮帮我，谢谢。

这是我的PHP代码：

 <？php  
  $ servername  =  本地主机; 
  $ username  =   root; 
  $ password  =  ; 
  //  创建连接 
  $ conn  =  new  mysqli（$ servername，$ username，$ password）; 
  //  检查连接 
  if （$ conn-> connect_error）
 {
 die（  Connection失败：。$ conn-> connect_error）; 
} 
  $ db_selected  = mysqli_select_db（$ conn，  test_json）; 
 if（！$ db_selected）
 {
 die（ Khôngthểsửdụng数据库： .mysql_error（））; 
} 
  $ sql  =   SELECT * from data; 
  $ result  =  $ conn   - >查询（$ SQL）; 
  $ outp  =   [; 
 while（$ rs = $ result-> fetch_array（MYSQLI_ASSOC））{
  if （$ outp！=   [）{$ outp。=  ，;} 
 $ outp。= '  {Name ：'。$ rs [ 名称]。' ，'; 
 $ outp。= ' 年龄：'。$ rs [ 年龄]。' ，; 
 $ outp。= ' 地址：'。$ rs [ 地址]。' ，; 
 $ outp。= ' Phone：'。$ rs [ 电话]。' }; 
} 
 $ outp。=  ]; 
 $ conn-> close（）; 
 echo（$ outp）; 
 ？>  

这是我的JSON

< pre lang =Javascript>< h1>客户< / h1 >
< div id = id01 > < / div >

< script>
var xmlhttp = new XMLHttpRequest（）;
var url = http：//本地主机：80 / demo.php;


xmlhttp.onreadystatechange = function （）
{
if （xmlhttp.readyState == 4 && xmlhttp.status == 200 ）{
myFunction（xmlhttp.responseText）;
}
}
xmlhttp.open（ GET， url， true ）;
xmlhttp.send（）;

function myFunction（response）{
var arr = < span class =code-sdkkeyword> JSON .parse（response）;
var i;
var out = < table> ;

for （i = 0 ; i< arr.length; i ++ ）{
out + = < tr>< td> +
arr [i] .Name +
< / td>< td> +
arr [i] .Age +
< / td>< td> +
arr [i] .Address +
< / td>< td> +
arr [i] .Phone +
< / TD>< / TR GT&;;
}
out + = < / table>
document .getElementById（ id01） .innerHTML = out;
}
< / script>

解决方案

servername = localhost;

用户名 = 根;

密码 = ;
// 创建连接

I try to reads JSON data from a web server running PHP and MySQL.
I learn code from http://www.w3schools.com/json/json_example.asp
And then I I changed their url with my urlm like this http://localhost/demo.php
But it does not work. I am beginner in JSON and I don't know what happened with my code. Please help me, thank.
Here is my PHP code:

<?php
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$conn = new mysqli($servername,$username,$password);
//check connection
if ($conn->connect_error) 
{
    die("Connection failed: " . $conn->connect_error);
}
$db_selected = mysqli_select_db($conn,"test_json");
if(!$db_selected)
{
	die("Không thể sử dụng DATABASE: ".mysql_error());
}
$sql = "SELECT * from data";
$result = $conn -> query($sql);
$outp = "[";
while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    if ($outp != "[") {$outp .= ",";}
    $outp.='{"Name":"'.$rs["Name"].'",';
    $outp.='"Age":"'.$rs["Age"].'",';
    $outp.='"Address":"'.$rs["Address"].'",';
	$outp.='"Phone":"'.$rs["Phone"].'"}';
}
$outp.="]";
$conn->close();
echo($outp);
?>

And here is my JSON

<h1>Customers</h1>
<div id="id01"></div>

<script>
var xmlhttp = new XMLHttpRequest();
var url = "http://localhost:80/demo.php";


xmlhttp.onreadystatechange=function() 
{
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        myFunction(xmlhttp.responseText);
    }
}
xmlhttp.open("GET", url, true);
xmlhttp.send();

function myFunction(response) {
    var arr = JSON.parse(response);
    var i;
    var out ="<table>";

    for(i = 0; i < arr.length; i++) {
        out +="<tr><td>" +
        arr[i].Name +
        "</td><td>" +
        arr[i].Age +
        "</td><td>" +
        arr[i].Address +
        "</td><td>"+
		arr[i].Phone +
        "</td></tr>";
    }
    out+="</table>"
    document.getElementById("id01").innerHTML = out;
}
</script>

解决方案

servername = "localhost";

username = "root";

password = ""; // Create connection

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