我们如何在PHP中传递2个值？

Question

我无法将所选值传递到列表页面。当我选择一个选项并单击搜索时，默认值将传递到列表页面。 （默认值：最后添加的值）

$sql = mysqli_query($con, "select * from jobregestation order by id");

while ($row = $sql->fetch_assoc()) {
    $id = $row['id'];
    $name = $row['JobName'];
    echo "<option value=" . $id . ">" . $name . "</option>";
}

提前致谢。

Answer 1

你期待这样的事吗？

<?php 

if(isset($_POST['submit'])){ // worked when click on serarch button

    echo $_POST['exampleName'];
}
?>
<form action="" method='post'>
<select name="exampleName">
<?php
   $sql = mysqli_query($con, "select * from jobregestation order by id");
   while ($row = $sql->fetch_assoc()) {
       $id = $row['id'];
       $name = $row['JobName'];
       echo "<option value=" . $id . ">" . $name . "</option>";
   }
?>
</select>
<input type="submit" name="submit" value="Search"/>

</form>

Answer 2

<select>
<?php
foreach ($sql->fetch_assoc() as $key=>$value)
{
echo "<option value = " .$value['id'] . ">" .$value['Jobname]."</option>";
}
?>
</select>

这应该有用。