Question

我正在尝试检查用户在SQL表上是否有特定字母，例如：

我有SQL表amx_amxadmins和列ID，用户名，访问示例：

amx_amxadmins:
--------------------------------------------------
| ID | USERNAME         | ACCESS                 |
--------------------------------------------------
| 1  | TEST1            | abcde                  |
--------------------------------------------------
| 2  | TEST2            | bcde                   |
--------------------------------------------------

标志A代表No Access。

那么，如何检查用户是否具有列访问权限A？

例如：

<?php
    $check_user_flag = $mysqli->query("SELECT * FROM `$amxadmins` WHERE `username` = '$logged[username]' AND `access` LIKE 'a' ") or die(mysqli_error($mysqli));
    $checkUSRFLAG = $check_user_flag->fetch_assoc();
    if($checkUSRFLAG->num_rows >  "a"){
        echo "test";
    }
    else {
        echo "test2";
    }
?>
```

RyffLe编辑

$query = "SELECT * FROM `$amxadmins` WHERE `username` = ?";
    $stmt = $mysqli->prepare($query);
    $stmt->bind_param('s', $param);
    $param = $logged['username'];
    $stmt->execute();
    $result = $stmt->bind_result();
    $stmt->close();

    $Permissions = $obj->access;

    if (strpos($Permissions, 'a')) {
       echo("An A has been found!");
    }
    else {
       echo("An A wasn't Found...");
    }

Answer 1

首先，我对Stackoverflow Beytula Ibryam表示欢迎。

那是您的桌子：

amx_amxadmins:
--------------------------------------------------
| ID | USERNAME         | ACCESS                 |
--------------------------------------------------
| 1  | TEST1            | abcde                  |
--------------------------------------------------
| 2  | TEST2            | bcde                   |
--------------------------------------------------

首先，您需要从数据中选择要检查的用户的所有数据：

$query = "SELECT * FROM `$amxadmins` WHERE `username` = ?";
$stmt = $mysqli->prepare($query);
$stmt->bind_param('s', $param);
$param = $logged[username];
$stmt->execute();
$result = $stmt->get_result();
$stmt->close();
$obj = $result->fetch_object();

下一步是获取您作为字符串的访问权限：

$Permissions = $obj->access;

现在，我们可以检查您想要的文字的访问权限：

if (strpos($Permissions, 'a')) {
   echo("An A has been found!");
}
else {
   echo("An A wasn't Found...");
}

享受;）

---编辑---

    $query = "SELECT * FROM `?` WHERE `username` = ?";
    $stmt = $mysqli->prepare($query);
    $stmt->bind_param('ss', $amxadmins, $logged['username']);
    $stmt->execute();
    $result = $stmt->get_result();
    $stmt->close();
    $obj = $result->fetch_object();

    if($checkUSRFLAG->num_rows >  "a"){
        echo ("test");
    }
    else {
        echo("test2");
    }

如果没有MySQLi连接文件，请在主文件中添加以下行：

$mysqli = new mysqli('YOUR_DATABEAS_IP', 'YOUR_DATABEAS_USER', 'YOUR_DATABEAS_USER_PASSWORD', 'YOUR_DATABEAS_NAME', 'YOUR_DATABEAS_PORT');
        if($mysqli->connect_error) {
          exit('Error connecting to database'); //Should be a message a typical user could understand in production
        }
        mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

Answer 2

听起来您只是在寻找通配符搜索。为此使用%字符。所以代替这个：

AND `access` LIKE 'a'

您将拥有：

AND `access` LIKE '%a%'

如何检查用户在表上是否存在特定字母

2 个答案: