Question

<br><h2>Select a Tag</h2></br>

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("portal", $con);

$query = "SELECT tag_name FROM tags";
$result = mysql_query($query);

?>
<select name="tag_name" id="abc">
<option size=30 selected>Select</option>
<?php
while($array = mysql_fetch_assoc($result)){
?>
<option value ="<?php echo $array['tag_name'];?>"><?php echo $array['tag_name'];?>      
</option>
 <?php

}
?>
</select>     <br><br>

这是用于获取页面中下拉菜单的代码片段。我有一个名为portal的数据库和名为tags的表，tag_name作为属性。帮我找到程序中的错误。我没有在下拉菜单中获取tag_names

Answer 1

</option>
 <?php
 <?php
}

以上是您的代码中的错误。

您重复了<?php

Answer 2

首先要做的事情：你能告诉我计数是否大于零？尝试使用“SELECT`tag_name` FROM tags”（使用奇怪的引号）而不是“SELECT tag_name FROM tags”来生成select语句。

Answer 3

<br><h2>Select a Tag</h2></br>

<?php

  // Connect to database
  // NEVER use the root user in a production environment!
  $con = mysql_connect("localhost","root","");
  if (!$con) {
    // NEVER show the result of mysql_error() in a production environment!
    die('Could not connect: ' . mysql_error());
  }
  mysql_select_db("portal", $con);

  // Run the query (and check the result)
  $query = "SELECT tag_name FROM tags";
  if (!$result = mysql_query($query)) {
    die("MySQL error at query: ".mysql_error());
  }

  if (mysql_num_rows($result)) {

    // Draw the select if the query returned more than 0 rows, or display an error

?>

<select name="tag_name" id="abc">
  <option size=30 selected>Select</option>

  <?php while ($array = mysql_fetch_assoc($result)) { ?>
  <option value="<?php echo $array['tag_name'];?>"><?php echo $array['tag_name'];?></option>
  <?php } ?>

</select><br><br>

<?php } else { ?>

Query returned 0 results!

<?php } ?>

PHP下拉菜单

3 个答案: