为什么 select 选项没有显示在 PHP 上
Why select option is not showing on PHP
<select class="gps_sel" name="gps_sel" id="gps_sel" style="color: #000;">
<option value="0">GPS Tracking</option>
<?php
require'connectDB.php';
$sql = "SELECT RTMGPS.gps_id
From RTMGPS
WHERE NOT EXISTS (SELECT employee_users.gps_id FROM employee_users WHERE RTMGPS.gps_id = employee_users.gps_id)";
$result= mysqli_query($conn,$sql);
$Arr= array();
while ($dataStat = mysqli_fetch_assoc($result))
{
$Arr[] = $dataStat;
}
?>
<option value="<?php json_encode($Arr) ?>"><?php echo $row['RTNumber']; ?></option>
<?php
?>
</select>
这是我的 select 选项下拉代码,但它没有显示我想要显示的数据
我认为这是正确的代码。你需要把选项放在 while 循环中。但实际上我不知道你为什么使用 json encode in value of that from,什么是 $row[''].
<select class="gps_sel" name="gps_sel" id="gps_sel" style="color: #000;">
<option value="0">GPS Tracking</option>
<?php
require'connectDB.php';
$sql = "SELECT RTMGPS.gps_id
From RTMGPS
WHERE NOT EXISTS (SELECT employee_users.gps_id FROM employee_users WHERE RTMGPS.gps_id = employee_users.gps_id)";
$result= mysqli_query($conn,$sql);
$Arr= array();
while ($dataStat = mysqli_fetch_assoc($result))
{
$Arr[]=json_decode($dataStat);
echo "<option value='".json_encode($Arr)."'>".$row['RTNumber']."</option>";
}
?>
</select>
<select class="gps_sel" name="gps_sel" id="gps_sel" style="color: #000;">
<option value="0">GPS Tracking</option>
<?php
require'connectDB.php';
$sql = "SELECT RTMGPS.gps_id
From RTMGPS
WHERE NOT EXISTS (SELECT employee_users.gps_id FROM employee_users WHERE RTMGPS.gps_id = employee_users.gps_id)";
$result= mysqli_query($conn,$sql);
$Arr= array();
while ($dataStat = mysqli_fetch_assoc($result))
{
$Arr[] = $dataStat;
}
?>
<option value="<?php json_encode($Arr) ?>"><?php echo $row['RTNumber']; ?></option>
<?php
?>
</select>
这是我的 select 选项下拉代码,但它没有显示我想要显示的数据
我认为这是正确的代码。你需要把选项放在 while 循环中。但实际上我不知道你为什么使用 json encode in value of that from,什么是 $row[''].
<select class="gps_sel" name="gps_sel" id="gps_sel" style="color: #000;">
<option value="0">GPS Tracking</option>
<?php
require'connectDB.php';
$sql = "SELECT RTMGPS.gps_id
From RTMGPS
WHERE NOT EXISTS (SELECT employee_users.gps_id FROM employee_users WHERE RTMGPS.gps_id = employee_users.gps_id)";
$result= mysqli_query($conn,$sql);
$Arr= array();
while ($dataStat = mysqli_fetch_assoc($result))
{
$Arr[]=json_decode($dataStat);
echo "<option value='".json_encode($Arr)."'>".$row['RTNumber']."</option>";
}
?>
</select>