Question

嘿，我是PHP / MySqli的新手，如果给定ID的Scanstatus字段已经是＆＃34;扫描＆＃34;，我想在更新前检查。如果它已经扫描显示消息为＆＃34;已经扫描＆＃34;否则更新。

以下代码仅更新，并且不检查是否已存在。

<?php 
$id = $_POST['id'];

$connection = mysqli_connect("localhost", "username", "passwd","dbname"); 
if(mysqli_connect_errno())
{
    echo "failed to connect " . mysqli_connect_error();
}

if(isset($_POST['Submit'])) 
{

    $query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id";
    $result = mysqli_query($connection,$query); 

    if (!$result) {
    die('Error' . mysqli_error($connection));
    }
    else
    {
    echo "Successfully updated";
    }
}
?>

Answer 1

尝试以下php代码

$id = $_POST['id'];

$connection = mysqli_connect("localhost", "username", "passwd","dbname"); 
if(mysqli_connect_errno())
{
echo "failed to connect " . mysqli_connect_error();
}

else{
if(isset($_POST['Submit'])) 
{
 $sql = "UPDATE sales SET ScanStatus = 'Scanned' WHERE id = '$id'";
      $result = $connection->query($sql);
    if ($result->num_rows > 0) {

              echo "Successfully updated";

      } else {
        die('Error' . mysqli_error($connection));
      }
}
}

单击SUBMIT按钮“isset（$ _ POST ['Submit']）”将为true且直接进入if语句。然后运行sql命令，如果有任何结果，sql查询受影响进入下一个if语句后，条件“$ result-＆gt; num_rows＆gt; 0”然后回显“成功更新”。

Answer 2

据我了解，您只想在ScanStatus字段不是Scanned

的情况下更新

因此，您可以修改现有的查询，而无需获取记录：

$query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id AND `ScanStatus` != 'Scanned'";

只需将查询更改为上方并使用：

if(mysqli_affected_rows($result) > 0 ){

这是完整的代码：

<?php 

$connection = mysqli_connect("localhost", "username", "passwd","dbname"); 
if(mysqli_connect_errno()) {
    echo "failed to connect " . mysqli_connect_error();
}

if(isset($_POST['Submit']))  {
    $id = $_POST['id'];
    $query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = $id AND `ScanStatus` != 'Scanned'";
    $result = mysqli_query($connection, $query);

    if(mysqli_affected_rows($connection) > 0 ){
        echo "Successfully updated";
    }
    else {
        echo 'Already Scanned';
    }
}

Answer 3

您可以使用此代码：

使用if(mysqli_affected_rows($mysqli) > 0 )或根本不进行比较。

用以下代码替换您的代码：

<?php 
$id = $_POST['id'];

$connection = mysqli_connect("localhost", "username", "passwd","dbname"); 
if(mysqli_connect_errno())
{
    echo "failed to connect " . mysqli_connect_error();
}

if(isset($_POST['Submit'])) 
{

    $my_query = mysqli_query($connection, "SELECT * FROM `sales` WHERE `id` = ". $id . " AND `ScanStatus` = 'Scanned'");
    if(mysqli_num_rows($my_query) > 0){
        echo "Already ScanStatus is Scanned";
    }
    else{
        $query = "UPDATE `sales` SET `ScanStatus` = 'Scanned' WHERE `id` = ".$id;
        //echo $query;die;
        $result = mysqli_query($connection, $query); 

        if(mysqli_affected_rows($connection) > 0 ){
           echo "Successfully updated";

           /* get new updated data */
           $new_query = mysqli_query($connection, "SELECT * FROM `sales` WHERE `id` = '$id' LIMIT 1");
           $new_info = mysqli_fetch_array($new_query);
           echo "<pre>"; print_r($new_info);

        }
        else
        {
            echo "Not updated";
        }
    }
}
?>

更新前PHP / Mysqli检查

3 个答案: