Question

嘿所有我试图使用if if / else语句来查询数据库并且我想要一个选项来从数据库中的特定列（在这种情况下是位置）中选择所有结果，另一个选项是仅根据表单选择项目输入：

<?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);
}

＆GT;

我很困惑的是selectlocation =''我不确定要放入什么，以便它返回所有结果。

任何帮助都会非常感激

Answer 1

你可以做到

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'"

OR

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

其中[xxx]是所选选项的值。

如何在查询数据库时使变量全部选中

1 个答案: