我有一个名为 user_id 的表中的存储值。我有另一个这样的表：
客户项目：
+------------------+-------------+-------------+-------------+
| project_codename | client_id_1 | client_id_2 | client_id_3 |
+------------------+-------------+-------------+-------------+
| Alpha | 1 | 2 | 3 |
+------------------+-------------+-------------+-------------+
| Beta | 2 | 1 | 0 |
+------------------+-------------+-------------+-------------+
| Gamma | 3 | 1 | 0 |
+------------------+-------------+-------------+-------------+
如果 user_id = (client_id_1 or client_id_2 or client_id_3) 那么我想在下拉列表中打印相应的 project_codenames。这是我拥有的 php，但没有返回正确的结果。我已经推销了我不确定的领域。
PHP（更新并解决）：
<?php
$con = mysqli_connect("localhost","****","****","db") or die("Connection error: " . mysqli_error($con));
//stored sample value from another table
$user_id = "1";
//Build query: Not sure if this is correct format
$query = "SELECT * FROM client_projects WHERE client_id_1 = $user_id OR client_id_2 = $user_id OR client_id_3 = $user_id";
$result = mysqli_query($con, $query) or die("Query error: " . mysqli_error($con));
//Drop down list
echo '<select id="Projects" class="input">';
echo '<option value="" selected="selected" disabled="disabled">Choose a Project...</option>';
// Loop through the query results, outputing specific options one by one
// Not sure of the loop for the options
while ($row = mysqli_fetch_array($query)) {
echo '<option value="'.$row['project_codename'].'">'.$row['project_codename'].'</option>';
}
echo '</select>';
mysqli_close($con);
?>