如何在查询数据库时使变量全选 - how to make a variable select all when querying a database

how to make a variable select all when querying a database

本文关键字：变量全选数据库查询 | 更新日期: 2023-09-27

嘿，我正在尝试使用if-if/else语句查询数据库，我想要一个选项从数据库中的特定列（在本例中为位置）选择所有结果，另一个选项仅根据表单输入选择项目：

<?php
if(isset($_POST['indexSearchSubmit']))
{
    if ($_POST['locationList']=='allLocations')
        {
            $selectedLocations = '*';
        }
    else
        {
            $selectedLocations = $_POST['locationList'];                
        }
    foreach($_POST['industryList'] as $selected)
        {
            $selectedIndustries = $selected;
            $result = mysqli_query($con,"SELECT * FROM currentListings WHERE location = '$selectedLocations' AND industry = '$selectedIndustries'");
            while($row = mysqli_fetch_array($result))
                {
                    echo $row['industry'];
                    echo "<br>";
                    echo $row['location'];
                    echo "<br>";
                }
        }
    mysqli_close($con);
}

？>

我感到困惑的是selectedlocation=''我不确定该放什么才能返回所有结果。

任何帮助都将不胜感激

您可以进行

if ($_POST['locationList']=='allLocations')
    {
        $l = 'IS NOT NULL';
    }
else
    {
        $l = " = '". $_POST['locationList'] . "'";
    }

和

$result = mysqli_query($con,"SELECT * FROM currentListings WHERE location " . $l . "  AND industry = '$selectedIndustries'");

这样，查询字符串将是

"SELECT * FROM currentListings WHERE location IS NOT NULL  AND industry = '$selectedIndustries'"

或

"SELECT * FROM currentListings WHERE location = '[xxx]'  AND industry = '$selectedIndustries'"

其中[xxx]是所选选项的值。