Question

我有4个输入并使用Ajax将数据发送到php文件：现在我很好奇如何将这些数据附加到json文件？

<input type="text" id="name">
<input type="text" id="surname">
<input type="text" id="mobile">
<input type="text" id="email">
<script>
var name = $("#name").val();
var surname = $("#surname").val();
var mobile = $("#mobile").val();
var email = $("#email").val();
$.ajax({type:"POST",
    url:"wjson.php",
    data:"name="+nombre+"&surname="+surname+"&mobile="+mobile+"&email="+email,
    success:function(data) {

    }
});

JSON文件：（people.json）

{
    "1":
    {
        "Name" : "Jhon",
        "Surname" : "Kenneth",
        "mobile" : 329129293,
        "email" : "jhon@gmail.com"
    },
    "2":
    {
        "Name" : "Thor",
        "Surname" : "zvalk",
        "mobile" : 349229293,
        "email" : "thor@gmail.com"
    }
}

wjson.php文件：

<?php
$nane = $_POST['name'];
$surname =$_POST['surname'];
$mobile = $_POST['mobile'];
$email =$_POST['email'];
$str_datos = file_get_contents("people.json")
//add new data to people.json
?>

通过people.json文件在我的服务器中的方式

Answer 1

// in your wjson.php file
$name = $_POST['name'];
$surname =$_POST['surname'];
$mobile = $_POST['mobile'];
$email = $_POST['email'];

$data = array(
    'Name'=>$name,'surname'=>$surname,
    'mobile'=>$mobile,'email'=>$email
);
$prevData = file_get_contents("people.json");
$prevData = json_decode($prevData,true);
$jsonData = json_encode(array_merge($prevData,array($data)));
file_put_contents('people.json',$jsonData);

Answer 2

你已经做了太多额外的工作

<?php
$str_datos = file_get_contents("people.json",true);
$str_datos[]=$_POST;
file_put_contents('people.json',json_encode($str_datos);
?>

Answer 3

将数据附加到json文件中试试这个。

<?php

if(file_exists("people.json")){

    $prevdata = file_get_contents("people.json");
    $newdata = $prevdata.','.json_encode($_POST);

    file_put_contents("people.json", $newdata);

}else{
    $data = json_encode($_POST);
    file_put_contents("people.json", $data);
}


?>

Answer 4

试试这个

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
  <form>
    <input type="text" id="name">
    <input type="text" id="surname">
    <input type="text" id="mobile">
    <input type="text" id="email">
    <input type="button" id="save" value="Save" />
  </form>
<script>
 $(document).ready(function(){
    $('#save').click(function(){
        var name = $("#name").val();
        var surname = $("#surname").val();
        var mobile = $("#mobile").val();
        var email = $("#email").val();
        $.ajax({
            type: "post",
            url: "wjson.php",
            data : {name : name, surname : surname, mobile: mobile, email : email },
            success:function(data) {
                console.log(data);
            },
            error:function(data){
                console.log(data);
            }
        });
    });
});
</script>

并用

替换wjson.php文件

if(file_exists('people.json'))
{
    $data = $_POST;
    $people = file_get_contents("people.json");
    $people = json_decode($people,true);
    $json = json_encode(array_merge($people,array($data)));
    file_put_contents('people.json',$json);
}
else
{
    $json = json_encode($_POST);
    file_put_contents('people.json',$json);
}

如何用php制作json文件

4 个答案: