选择、插入和更新 MySQL 表 PHP 脚本不起作用答案

【问题标题】：Select, Insert and Update MySQL table PHP script not working选择、插入和更新 MySQL 表 PHP 脚本不起作用
【发布时间】：2015-08-17 13:46:57
【问题描述】：

我写了以下代码：

<?php
    $servername = "domain";
    $insert = 12345678;
    $username = "user";
    $password = "password";
    $dbname = "database";

// Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $sql = "SELECT quantity FROM Eshop WHERE id = $insert";
    $result = $conn->query($sql);
    echo $result;
    if ($result > 0) {
        $result = $result + 1;
        $sql2 = "UPDATE Eshop SET quantity = $result WHERE id = $insert";
        if ($conn->query($sql2) === TRUE) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql2 . "<br>" . $conn->error;
        } 
    } else {
        $sql3 = "INSERT INTO Eshop (id, quantity) VALUES ($insert, 1)";
        if ($conn->query($sql3) === TRUE) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql3 . "<br>" . $conn->error;
        }
    }
?>

我想要这个脚本做的是选择quantity，其中id 是$insert，然后如果quantity > 0，将1 添加到quantity 并更新quantity，如果没有插入id 和 quantity = 1 到表中。该表只有两个字段id(VARCHAR(32)) 和quantity(DECIMAL(8,1))。为了帮助尽可能多的人，我尝试将其做得更笼统。请你帮助我好吗？提前致谢。注意：当我在浏览器中运行脚本时（在使用正确的用户名、域等将其上传到服务器之后）什么都没有显示，我什至在控制台中都没有收到错误。

【问题讨论】：

用引号将您的 php 变量包含在查询中，例如：“SELECT quantity FROM Eshop WHERE id = '$insert'”;
感谢您的帮助，但很遗憾没有成功。

标签： php mysql select insert sql-update

【解决方案1】：

    <?php
    $servername = "domain";
    $insert = 12345678;
    $username = "user";
    $password = "password";
    $dbname = "database";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $sql = "SELECT quantity FROM Eshop WHERE id = $insert";
    $result = $conn->query($sql);

    //check if particular record exists or not
    $count=mysql_num_rows($result);

   if($count>0) // if records exists for the particular id
   {
    while($row = mysqli_fetch_array($result)) {
        if ($row['quantity'] > 0) {
            $new_quantity = $row['quantity'] + 1;
            $update = "UPDATE Eshop SET quantity = '$new_quantity' WHERE id = '$insert'";
            if ($conn->query($update ) == TRUE) {
                echo "New record created successfully";
            } else {
                echo "Error: " . $update . "<br>" . $conn->error;
            } 
        } else {
            $insert = "INSERT INTO Eshop (id, quantity) VALUES ('$insert', 1)";
            if ($conn->query($insert ) === TRUE) {
                echo "New record created successfully";
            } else {
                echo "Error: " . $insert . "<br>" . $conn->error;
            }
        } // else qty is <=0

    } //end while
}
else {
   echo "Records do not exists for that particular id";
 }
?>

【讨论】：

【解决方案2】：

您需要提取查询结果并遍历每一行：

<?php
    $servername = "domain";
    $insert = 12345678;
    $username = "user";
    $password = "password";
    $dbname = "database";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $sql = "SELECT quantity FROM Eshop WHERE id = $insert";
    $result = $conn->query($sql);

    while($row = mysqli_fetch_array($result)) {
        if ($row['quantity'] > 0) {
            $new_quantity = $row['quantity'] + 1;
            $sql2 = "UPDATE Eshop SET quantity = '$new_quantity' WHERE id = '$insert'";
            if ($conn->query($sql2) == TRUE) {
                echo "New record created successfully";
            } else {
                echo "Error: " . $sql2 . "<br>" . $conn->error;
            } 
        } else {
            $sql3 = "INSERT INTO Eshop (id, quantity) VALUES ('$insert', 1)";
            if ($conn->query($sql3) == TRUE) {
                echo "New record created successfully";
            } else {
                echo "Error: " . $sql3 . "<br>" . $conn->error;
            }
        }

    }
?>

【讨论】：