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[英]php drop down list with “two values” for each “option” and multiple selections
[英]Set the value for each option of an HTML drop down list to a PHP variable
我希望我的下拉列表在選項中顯示sql表中的名稱,但值應為sql表中的ID。 我該如何實現?
<select name="Warehouse">
<?php
$servername = "127.0.0.1";
$username = "root";
$password = "12345";
$db_name = "Second";
$the_port = "3306";
// Create connection
$conn = new mysqli($servername, $username, $password,$db_name,$the_port);
$conn->set_charset('utf8');
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
echo "err";
}
// fetch ID,Name_of from table
$sql="SELECT ID,Name_of FROM WAREHOUSE;";
$result = $conn->query($sql);
while($row = mysqli_fetch_assoc($result)) {
$val=$row['ID'];
echo "<option value=".$val." >" . $row['Name_of'] . "</option>";
}
?></select>
實際上錯誤是在sql查詢中,在SELECT中我從未獲得ID。
不管怎么說,還是要謝謝你。
您的回聲代碼需要一些更改,它應該像這樣
echo "<option value='".$val."' >" . $row['Name_of'] . "</option>";
^^^ ^^^
您需要將查詢更改為
$sql="SELECT ID,Name_of FROM WAREHOUSE";
更改您的查詢表格
$sql="SELECT Name_of FROM WAREHOUSE;";
至
$sql="SELECT ID,Name_of FROM WAREHOUSE;";
您可以嘗試以下方法:
<option value="$val">$row['Name_of']</option>
這樣嘗試
echo "<option value=".$row['ID']." >" . $row['Name_of'] . "</option>";
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