php - 如何在 INSERT INTO PHP 中使用字符串变量作为表目标

Question

我有这个代码

<?php
$servername = "localhost";
$username = "username";
$password = "pass";

// Create connection
$conn = new mysqli($servername, $username, $password);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

//Take the data from the request
$json = file_get_contents('php://input');
$data = json_decode($json, true);

$tableto = $data['tabledestination'];
$Question = $data['question'];
$answerone = $data['answerone'];
$answertwo = $data['answertwo'];
$answerthree = $data['answerthree'];
$Feedback = $data['feedback'];
$Correctanswer = $data['Correctanswer'];
$reparcial = $data['partial'];


//Insert the data into the database


$sql = "INSERT INTO mydatabase."$tableto" (Question, Answer_1, Answer_2, Answer_3, Correct_Answer, Parcial, Feedback)
VALUES ('" . $Question . "', '" . $answerone . "', '" . $answertwo . "', '" . $answerthree . "', '" . $Correctanswer . "', '" . $reparcial . "', '" . $Feedback . "')";

if ($conn->query($sql) === TRUE) {
    echo '{ "result": "success" }';
} else {
    echo '{ "result": "error" }';
}


//Close the connection
$conn->close();

?>

我通过带有 HTTPMethod“POST”的 URLRequest 传递信息我想通过我发布的字典发送我的数据库的表目标，这样我就可以根据我传递的字符串分配表的目标

Answer 1

以下

$sql = "INSERT INTO mydatabase."$tableto" (Question, Answer_1, Answer_2, Answer_3, Correct_Answer, Parcial, Feedback)
VALUES ('" . $Question . "', '" . $answerone . "', '" . $answertwo . "', '" . $answerthree . "', '" . $Correctanswer . "', '" . $reparcial . "', '" . $Feedback . "')";

因为使用"

而出现问题

当声明 $sql 时，您正在使用字符串的连接，而您错过了 .在你的连接中。

尝试更改为:

$sql = "INSERT INTO mydatabase." . $tableto . " (Question, Answer_1, Answer_2, Answer_3, Correct_Answer, Parcial, Feedback)
    VALUES ('" . $Question . "', '" . $answerone . "', '" . $answertwo . "', '" . $answerthree . "', '" . $Correctanswer . "', '" . $reparcial . "', '" . $Feedback . "')";