$i = 1;
$sql = "
SELECT
heroes.char_id, characters.char_name, heroes.class_id, heroes.count,
heroes.played, heroes.active FROM heroes, characters
WHERE characters.obj_Id = heroes.char_id AND heroes.played = 1
";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query))
{
echo "
<tr>
<td><span id='lefttop'><b><font color='#007aa2'>".$i++." </font></td><td><b><font color='#f6ff00'>".$row['char_name']."</font></td></span><div style='float:right;'><td><b> ".$row['count']."</td></div> <br />
</tr>
";
}
echo "";
?>有人能帮上忙吗?我正试图在java服务器上创建一个英雄状态脚本,并且我需要连接到两个表,即“hero”(这是我获得英雄的地方)和"character“(这是我获得英雄名字的地方)。
发布于 2013-10-09 17:05:02
试试这个Temple:
<?php
$sql1 = "SELECT * your table1";
$query1 = mysql_query($sql);
while($row1 = mysql_fetch_array($query1))
{
echo "<tr>";
$sql2 = "SELECT * your table2";
$query2 = mysql_query($sql2);
while($row2 = mysql_fetch_array($query2))
{
echo "<td>[yor Vars here]</td>";
}
echo "</tr>";
}
?>https://stackoverflow.com/questions/19265141
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