我使用以下PHP代码从mysql db table team_details
检索团队名称,其中包含2列,team_id
和team_name
<?php
mysql_connect("", "", "");
mysql_select_db("db_name");
$data = mysql_query("SELECT team_name FROM team_details");
print "Team A:";
Print "<select name="dropdown">";
while($info = mysql_fetch_array( $data ))
{
Print "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
Print '</select>';
?>
完整代码
VS
Team B:
<select id="team2" disabled="true">
<option value="volvo">Volvo</option>
<option value="saab">Saab</option>
<option value="opel">Opel</option>
<option value="audi">Audi</option>
</select>
<button id="sub" disabled="true" name ="choose" onclick="load_players();"> Choose Players</button>
<button type="reset" value="Clear" onclick="reset_load();"> Clear</button>
</form>
</div>
</body>
答案 0 :(得分:1)
你的错误可能在这里:
Print "<select name="dropdown">";
你需要逃避双引号:
Print "<select name=\"dropdown\">";
答案 1 :(得分:0)
首先,这里有语法错误:
Print "<select name="dropdown">";
这样的逃脱语录:
Print "<select name='dropdown'>";
其次,使用mysql_fetch_assoc而不是mysql_fetch_array,因此您的代码如下所示:
while($info = mysql_fetch_assoc( $data ))
{
Print "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
检查你的mysql_connect以获得正确的值(主机名,用户,密码)
答案 2 :(得分:0)
尝试以下代码,替换host,db_username,db_password和db_name
mysql_connect("host", "db_username", "db_password");
mysql_select_db("db_name");
$data = mysql_query("SELECT team_name FROM team_details");
echo "Team A:";
echo "<select name='dropdown'>";
while($info = mysql_fetch_array( $data ))
{
echo "<option value='".$info['team_name']."'>".$info['team_name'] . "</option> ";
}
echo '</select>';