Question

我有两个php文件，文件A和文件B，它们从同一数据库中获取数据。文件A在站点的段落中显示数据，而文件A在AJAX中使用文件B在用graph.js创建的图形中显示数据。除了这两个文件，它们都可以工作。

在两个文件中，我都有一个名为$employeeNumber的变量，它是输入字段的名称，该格式由用户用来输入他们想查看其数据的任何员工编号。文件A允许我使用 $employeeNumber = $_POST["employeeNumber"];获取输入并准确显示数据，但是文件B不允许我使用$_POST[]，$_REQUEST[]或$_GET[]。相反，它迫使我对唯一编号进行硬编码。这是不希望的。

文件B用来将输入输入到表单中，并使用与输入的员工编号有关的数据更新图形。

我对文件A使用了回显$row["empnumber"];来显示数据，对于文件B我使用了print json_encode($data);了这是我出了错的地方吗？

文件B（除非我对employeeNum进行硬编码，否则不会更新图形）

<?php

header('Content-Type: application/json');

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";




//BELOW IS WHERE THE ISSUE LIES.
//WHEN I HARDCODE THE UNIQUE NUMBER LIKE THIS, THE GRAPH UPDATES ACCORDINGLY. WITH ACCURATE DATA FROM THE EMPLOYEE THAT THE UNIQUE NUMBER BELONGS TO.

$employeeNum = 8020;


//THIS (BELOW) I ANTICIPATED WOULD GET THE EMPLOYEE NUMBER FROM THE USERS INPUT IN THE FORM WHICH I HAVE NOW INCLUDED BUT IT PRODUCES AN ERROR. 

$employeeNum = $_POST["employeeNum"]; //this is line 9

//HERE IS THE ERROR WHEN I OPEN THE Graphdata.php FILE

<br />
<b>Notice</b>:  Undefined index: employeeNum in <b>C:\wamp64\www\Graphdata.php</b> on line <b>9</b><br />
<br />
<b>Warning</b>:  Invalid argument supplied for foreach() in <b>C:\wamp64\www\Graphdata.php</b> on line <b>22</b><br />
<br />
<b>Fatal error</b>:  Uncaught Error: Call to a member function close() on bool in C:\wamp64\www\Graphdata.php:26
Stack trace:
#0 {main}
  thrown in <b>C:\wamp64\www\Graphdata.php</b> on line <b>26</b><br />




//Connect to database
$dbconnect = new mysqli($servername, $username, $password, $dbname);

//If failed to connect
if(mysqli_connect_error()){
    die("Database connection error".mysqli_connect_error());
}
//Create query
$sql = "SELECT Mathematics, English, Afrikaans, Geography, Physics, History, Life_Orientation FROM studentresults WHERE employeeNumber = $employeeNum";

//Run query
$result = $dbconnect->query($sql);

//Produce result
$data = array();
foreach ($result as $row){ //THIS IS LINE 22
    $data[] = $row;
}

$result -> close();

$dbconnect -> close();

//Print the result
print json_encode($data);
?>

文件A（工作正常）



<?php 

//connecting to database
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "mydb";
    $employeeNum = $_POST["employeeNum"];

    $connect = new mysqli($servername, $username, $password, $dbname);

//checking connection

    if (mysqli_connect_error()) {
        die ("Database connection failed: ". mysqli_connect_error());
    } 

// Query the database
        $sql = "SELECT Salary FROM myEmployees WHERE employeeNumber = $employeeNum";

        $result = $connect->query($sql);

//print out result
while($row = mysqli_fetch_array($result)) {
        echo $row["English"];
    }

$result -> close();
?>

这是发出请求的AJAX代码

$.ajax({

//Graphdata.php is the name of FILE B

    url: "http://localhost/Graphdata.php",
    type: "GET",
    success : function(data){
        console.log(data);

        var Salary = [];
        var timeWorked = [];
        var Age = [];
        var userid = [];

        for(var i in data){

            userid.push("UserID " + data[i].userid);
            Salary.push(data[i].Salary);
            Age.push(data[i].Age);
            timeWorked.push(data[i].timeWorked);


        }
//Create the graph
        var ctx = document.getElementById('myChart').getContext('2d');

var chart = new Chart(ctx, {

这是表格代码。

<form action="Graphdata.php" method="post">
    <p class="name">Username</p>
<input class="inp1" type="text" name="name">
    <p class="password">Password</p>
<input class="inp2" type="password" name="email">
    <p class="employeeid">Employee ID</p>
<input id = "employeeNum" class="inp3" type="int" name="employeeNum">

Answer 1

问题在于，当您通过AJAX调用时，您没有将任何数据传递到后端。

此外，当您使用method=GET时，$_POST数组将为空。

将您的方法更改为POST，并包含如下数据：

$.ajax({
  method: "POST",
  url: "http://localhost/Graphdata.php",
  data: {employeeNum: someValue}
  ...
})

如何在Json打印php文件中将表单输入设置为变量？

1 个答案: