我在 php 中有一個來自數據庫的選擇下拉列表。如果該值已存在于列中，則不應顯示在選擇下拉列表中，因此我編寫了以下代碼：<select class="form-control" id="space" name="space"> <option value="--Select--">--Select--</option> <?php $select=mysqli_query($con,"select * from clients"); while($menuz=mysqli_fetch_array($select)) { $filled =$menuz['Space']; $valuez = array("C101","C102","C103","C104","C105","C106","C107","C108","W1","W2","W3","W4","W5","W6","W7","W8","W9","W10","W11","W12","F1","F2","F3","F4","F5","F6","F7","F8","F9","F10"); foreach($valuez as $value){ if($value != $filled){ ?> <option value="<?php echo $value;?>"> <?php echo $value; ?> </option> <?php } }}?> </select>現在的問題是值在選擇列中顯示了兩次。第一次顯示所有值，然后顯示我想要的不在數據庫中的值。誰能幫我解決這個問題。
1 回答
MYYA
TA貢獻1868條經驗 獲得超4個贊
看看以下選項：
選項 1：剛剛在 SQL 查詢中添加了 DISTINCT
<select class="form-control" id="space" name="space">
<option value="--Select--">--Select--</option>
<?php
$select=mysqli_query($con,"select DISTINCT `Space` from clients");
while($menuz=mysqli_fetch_array($select)){
$filled =$menuz['Space'];
$valuez = array("C101","C102","C103","C104","C105","C106","C107","C108","W1","W2","W3","W4","W5","W6","W7","W8","W9","W10","W11","W12","F1","F2","F3","F4","F5","F6","F7","F8","F9","F10");
foreach($valuez as $value){
if($value != $filled){
?>
<option value="<?php echo $value;?>">
<?php echo $value; ?>
</option>
<?php
}
}
}
?>
</select>
或者
選項 2：僅使用“NOT IN”忽略 SQL 查詢中的現有值集。在這里，您可以使用準備好的語句來避免 SQL 注入。僅出于演示目的，我在這里展示了一個帶有嵌入式輸入參數的查詢。
<?php
$valuez = array("C101","C102","C103","C104","C105","C106","C107","C108","W1","W2","W3","W4","W5","W6","W7","W8","W9","W10","W11","W12","F1","F2","F3","F4","F5","F6","F7","F8","F9","F10");
?>
<select class="form-control" id="space" name="space">
<option value="--Select--">--Select--</option>
<?php
$select=mysqli_query($con,"SELECT DISTINCT `Space` FROM `clients` WHERE `Space` NOT IN ('". implode("', '", $valuez) ."')");
while($menuz=mysqli_fetch_array($select)){
$filled =$menuz['Space'];
?>
<option value="<?php echo $filled;?>">
<?php echo $filled; ?>
</option>
<?php
}
?>
</select>
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