Question

我一直在努力警告：mysql_fetch_row（）期望参数1是资源，在第6行的xx中给出布尔值

错误此代码有什么问题？我该如何解决？

$read = mysql_query("select * from detail");

while($wr = mysql_fetch_array($read)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}

编辑;我这样做仍然是错误的。

$db = new mysqli('localhost', 'root', '', 'panel');
$sql = "select * from detail";
$read = $db->query($sql);

while($wr = mysql_fetch_array($read)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}

Answer 1

您正在混合MySQL和MySQLi扩展！
使用其中任何一个的appropriate functions，而不是两者。

Answer 2

由于查询失败，您可能在查询数据库时遇到问题：

$read = mysql_query("select * from detail"); // $read is false

// You can try to discover the error.
$read = mysql_query("select * from detail") || die(mysql_error());

Answer 3

使用此代码：

$read = mysql_query("select * from detail");

while($wr = mysql_fetch_array($read,  MYSQL_ASSOC)) {
echo $wr['Who'];
echo "<br />";
echo $wr['Time'];
echo "<br />";
echo $wr['What'];
}

有关详细信息，请参阅：mysql-fetch-array

用php从数据库中提取数据

3 个答案: