发送Javascript信息到PHP并保存到数据库中

Question

一位前同事开发了一个测验，它将结果发送到数据库。他被解雇了，但我需要再次使用该代码。我只有javascript代码，我需要重新创建保存来自javascript的信息的php（save.php）。你可以帮助我的PHP代码或给我一个提示。 Thaks！

$(document).ready(function() { 

    $("#answer_a").click(function() { 
    $.get("http://nameOfWebsite/save.php", {test: "1", question: "1", answer: "a" }); 
}); 

    $("#answer_b").click(function() { 
    $.get("http://nameOfWebsite/save.php", {test: "1", question: "1", answer: "b" }); 
}); 

    $("#answer_c").click(function() { 
    $.get("http://nameOfWebsite/save.php", {test: "1", question: "1", answer: "c" }); 
}); 

    $("#answer_d").click(function() { 
    $.get("http://nameOfWebsite/save.php", {test: "1", question: "1", answer: "d" }); 
}); 


    }); 

Answer 1

在save.php使用$ _GET []以使用变量，并将它们保存在你的桌子

<?php 
$con=mysqli_connect("example.com","peter","abc123","my_db"); 
// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

// escape variables for security 
$test = mysqli_real_escape_string($con, $_GET['test']); 
$question = mysqli_real_escape_string($con, $_GET['question']); 
$answer = mysqli_real_escape_string($con, $_GET['answer']); 

$sql="INSERT INTO Persons (test, question, answer) 
VALUES ('$test', '$question', '$answer')"; 

if (!mysqli_query($con,$sql)) { 
    die('Error: ' . mysqli_error($con)); 
} 
echo "1 record added"; 

mysqli_close($con); 
?> 

Answer 2

使用$_GET[<name>]让你的PHP的价值观和mysqli_connect将数据插入到桌子。

Answer 3

您已经走上正轨，下一步就是在PHP中。你可以使用这个例子来获取这些值。考虑这个例子：

<?php 
if(isset($_GET['test'])) { 
    $data = array(); // initialize return data holder 
    $test = isset($_GET['test']) ? $_GET['test'] : null; 
    $question = isset($_GET['question']) ? $_GET['question'] : null; 
    $answer = isset($_GET['answer']) ? $_GET['answer'] : null; 
    // they should be inside now, now you can go on with mysql inserts 

    // just a sample callback value to check if indeed php got it 
    $data['test'] = $test; 
    $data['question'] = $question; 
    $data['answer'] = $test; 

    echo json_encode($data); 
    exit; 
} 


?> 

<!-- lets say this is an image --> 
<button id="answer_a" type="button">Hi im an image</button> 

<script src="jquery.min.js"></script> 
<script type="text/javascript"> 
$(document).ready(function(){ 

    $("#answer_a").click(function() { 
     $.get("index.php", {test: "1", question: "1", answer: "a" }, function(response){ 
      var data = $.parseJSON(response); 
      console.log(data); // check this in console 
     }); 
    }); 

}); 
</script>