使用PHP更新MySQL數據庫

Question

嗨，我正在嘗試使用php更新我的mysql數據庫。 我可以用以下內容完美地更新它：

<?php

$conn = mysqli_connect("localhost", "root", "", "logintest");

if(!$conn){
    die("Connection failed: ".mysqli_connect_error());
}
?>

<?php

    $sql = "UPDATE user SET bot = '1' WHERE id = 9";

    if($conn -> query ($sql) === TRUE){
        echo "record updated successfully";
    }else{
        echo "Error updating record" . $conn -> error;
    }

    $conn -> close ();

?>

但是在將上面的bot列更新為1之前，我想檢查一下它是否為0，因為它只能是0或1。為此，我執行了以下操作（請參見下文），但它不起作用，是可能，或者是否有其他方法？ 感謝所有幫助，謝謝！！

$sql = "SELECT bot FROM user"; // bot is the column in the table which should be 0 or 1

        if( $sql == '0') { //if its 0 i can update it

            echo 'here'; //if i get here i will update using code above
     }

Answer 1

有兩種方法

1. SELECT和UPDATE

    $query = "SELECT bot FROM user where id=9" $res = $conn->query($query); if ($res->num_rows == 1) { // it should return only one row as id is unique $row = $result->fetch_assoc() if($row["bot"] == 0){ // UPDATE } } 

2. CASE構造

    UPDATE user SET bot = CASE WHEN bot = 0 THEN 1 ELSE bot END WHERE id='9' 

這個怎么運作：

它會根據匹配案例的返回值來更新機器人值。 如果當前bot值為0 ，則返回1 ，否則返回id=9行的current value 。

優勢：僅1個查詢

Answer 2

請試試

<?php

$conn = mysqli_connect("localhost", "root", "", "logintest");

if(!$conn){
    die("Connection failed: ".mysqli_connect_error());
}

$sql= "SELECT bot FROM user where columnid=value"; // change is according to your real value
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        if($row["bot"] == 0){
            $sql = "UPDATE user SET bot = '1' WHERE id = 9";

            if($conn -> query ($sql) === TRUE){
                echo "record updated successfully";
            }else{
                echo "Error updating record" . $conn -> error;
            }
        }
    }
}

$conn -> close ();
?>