php - MySQL 未在表中找到特定行

Question

这个问题已经有答案了:

When to use single quotes, double quotes, and backticks in MySQL (13 个回答)

Are table names in MySQL case sensitive? (5 个回答)

已关闭 4 年前。

我刚刚接触 PHP，我想制作一个简单的数据查找脚本。这是我现在在 index.php 文件中使用的代码。

<?php
    $mysqlHost = 'localhost';
    $mysqlUser = 'root';
    $mysqlPass = 'password';
    $mysqlDatabase = 'core';

    $conn = new mysqli($mysqlHost, $mysqlUser, $mysqlPass, $mysqlDatabase);
    if ($conn -> connect_error)
        die('Connection Failed: ' . $conn -> connect_error);

    $sql = 'SELECT * FROM users WHERE name = SimonM34';
    $result = $conn -> query($sql);

    if ($result -> num_rows > 0) {
        $row = $result -> fetch_assoc();
        $rawJson -> name = $row['name'];
        $rawJson -> rank = $row['rank'];
        $rawJson -> subrank = $row['subrank'];
        $rawJson -> balance = $row['balance'];
        $json = json_encode($rawJson);
        echo $json;
    } else {
        echo 'No users found!';
    }

    $conn -> close();

?>

这总是显示 No users found但如果我将语句更改为 SELECT * FROM users它返回{"name":"SimonM34","rank":"Guest","subrank":"None","balance":"500"}这很奇怪，我不明白为什么会发生这种情况。

谢谢

Answer 1

问题出在您的查询中:

$sql = 'SELECT * FROM users WHERE name = SimonM34';

您正在比较不带引号的字符串值。将您的查询替换为如下内容:

$sql = 'SELECT * FROM users WHERE name = "SimonM34"';

这是我认为唯一的问题。