PHP mysql中的多个选择条件 - multiple select conditions in php mysqli

multiple select conditions in php mysqli

本文关键字：选择条件 mysql PHP | 更新日期: 2023-09-27

我有一个像上面那样的表

我想选择和显示成本时，拾取和放置是相同的，由用户从下拉菜单选择具有以上选项。

$result = mysqli_query($conn,"SELECT * FROM location
WHERE Pickup='". $pick. "' ");
while($row = mysqli_fetch_array($result))
{
    echo "You have to pay " . $row['Cost'];
    echo "<br>";
}

上面的代码将给我4个选项，因为我没有检查dropLocation的等式。

SELECT * FROM location
WHERE Pickup='". $pick. "' AND DropLocation='". $drop. "'  "

它不工作。如何处理多个条件?

试试这个

你的代码看起来有SQL攻击漏洞尝试使用mysqli准备语句。

            <?php
            $stmt = $conn->prepare("SELECT * FROM location WHERE Pickup=? AND DropLocation=?");
            $stmt->bind_param('ss',$pick,$drop);
            //The argument may be one of four types:
            //i - integer
            //d - double
            //s - string
            //b - BLOB
            //change it by respectively 
            $stmt->execute();
            $get_result =$stmt->get_result();
            $row_count= $get_result->num_rows;
            if($row_count>0)
            {
                    while($row = mysqli_fetch_array($get_result))
                    {
                        echo "You have to pay " . $row['Cost'];
                        echo "<br>";
                    }

            }
             else
             {
                 echo "result set empty";
             }

             $stmt->close();
             $conn->close();
            ?>

试试这个:

if(isset($_POST['myPick']) && !isset($_POST['myDrop'])){
$result = mysqli_query($conn,"SELECT * FROM location
WHERE Pickup='". $pick. "' ");
}else if(!isset($_POST['myDrop']) && isset($_POST['myPick'])){
$result = mysqli_query($conn,"SELECT * FROM location
WHERE DropLocation='". $drop. "' ");
}else if(isset($_POST['myDrop']) && isset($_POST['myPick'])){
$result = mysqli_query($conn,"SELECT * FROM location
WHERE Pickup='". $pick. "' AND DropLocation='". $drop. "' ");
}
while($row = mysqli_fetch_array($result))
{
    echo "You have to pay " . $row['Cost'];
    echo "<br>";
}