我有这段代码。出于某种原因,当我显示html时,它不会显示id列的值。mysql表中有三列:"id"、"username"answers"password"。有人能帮我吗?
include "Connector.php";
$query = "SELECT id, username, password FROM users";
$result = mysql_query("SELECT username, password FROM users") or die ("MySQL error: " .mysql_error());
echo "<table>";
while($row = mysql_fetch_array($result)){
echo "
<tr>
<td>"
.$row['id'].
"</td>
<td>"
.$row['username'].
"</td>
<td>"
.$row['password'].
"</td>";
echo "<td><a href='deleteRow.php?oId=".$row[password]."'>Delete</a></form></td>
</tr>";
}
echo "</table>";
mysql_free_result($result);
mysql_close($connection);
您不选择id列:
$query = "SELECT id, username, password FROM users";
$result = mysql_query("SELECT username, password FROM users")...
mysqlquery行中缺少id列。$query变量根本没有使用。让我猜猜。。。你把代码复制粘贴了吗?
这样做:
$result = mysql_query($query)....
$result = mysql_query("SELECT username, password FROM users") or die ("MySQL error: " .mysql_error());
^^^^^^^^^^^^^^^^^^^^
没有id
字段。
您有两个查询,其中一个实际上正在运行。这是实际运行的查询:
$result = mysql_query("SELECT username, password FROM users")
您需要将"id"添加到SELECT中,或者只使用上面的$query变量。
使用此查询查找所有列的值:
$result=mysql_query("SELECT id,username,password FROM users")或die("mysql错误:".mysql_error());