SAE中查询数据返回异常警告
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2022-04-23 12:13:27
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SAE中查询数据返回错误警告
SAE平台，数据库已开启，表格已经建好，希望实现功能：查询字段A包含内容a的记录，返回该记录的字段B值：
代码如下：
$mysql = new SaeMysql();
echo "Connect success
\n";
$sql = mysql_query("SELECT * FROM `nature` WHERE `name` LIKE \'狼\' LIMIT 0, 30 ");
$data = $mysql->getData($sql);
echo $data['class'];
?>
运行结果为：
Connect success
Warning: mysql_query() [function.mysql-query]: this app is not authorised in index.php on line 5
Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in index.php on line 5
Warning: mysqli_query() [function.mysqli-query]: Empty query in saemysql.class.php on line 191
请问下各位大大，是错在哪呢？
------解决思路----------------------
$sql = "SELECT * FROM `nature` WHERE `name` LIKE '%狼%' LIMIT 0, 30 ";
SAE平台，数据库已开启，表格已经建好，希望实现功能：查询字段A包含内容a的记录，返回该记录的字段B值：
代码如下：
$mysql = new SaeMysql();
echo "Connect success
\n";
$sql = mysql_query("SELECT * FROM `nature` WHERE `name` LIKE \'狼\' LIMIT 0, 30 ");
$data = $mysql->getData($sql);
echo $data['class'];
?>
运行结果为：
Connect success
Warning: mysql_query() [function.mysql-query]: this app is not authorised in index.php on line 5
Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in index.php on line 5
Warning: mysqli_query() [function.mysqli-query]: Empty query in saemysql.class.php on line 191
请问下各位大大，是错在哪呢？
------解决思路----------------------
$sql = "SELECT * FROM `nature` WHERE `name` LIKE '%狼%' LIMIT 0, 30 ";
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