[英]having problem with mysqli_query, return NULL but the query works in phpmyadmin
当用户从iOS应用发送登录命令时,我正试图从数据库中检索信息。 要测试此功能,我将手动启动php页面(例如http://www.testdatabase.com/LoginFunctions.php )并以编程方式强制使用用户名。
问题是mysqli_query返回NULL值。 如果我使用“ or die(mysql_error()”),则不会发生任何事情。即使我使用mysqli_num_rows返回1,但$ result仍为空。因此,当执行mysql_fetch_assoc时,程序崩溃而未显示任何错误。
<?php
// Create connection
$con=mysqli_connect("localhost","super","super","testdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$action = "login";
$username = "Peperoncino";
$response = array();
if ($action == "login")
{
$query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";
if ($result = mysqli_query($con, $query))
{
$values = mysql_fetch_assoc($result);
$password = $values['pswrd'];
$response["password"] = $password;
$response["message"] = "Get information from db";
}
else
{
echo "err";
}
echo json_encode($response);
}
// Close connections
mysqli_close($con);
?>
您正在使用不推荐使用的mysql_fetch函数。使用新的
<?php
// Create connection
$con=mysqli_connect("localhost","super","super","testdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$action = "login";
$username = "Peperoncino";
$response = array();
if ($action == "login")
{
$query = "SELECT psw AS pswrd,id FROM Activities WHERE nome = 'Peperoncino' LIMIT 1";
if ($result = mysqli_query($con, $query))
{
$values = mysqli_fetch_assoc($result);
$password = $values['pswrd'];
$response["password"] = $password;
$response["message"] = "Get information from db";
}
else
{
echo mysqli_error($conn);
}
echo json_encode($response);
}
// Close connections
mysqli_close($con);
?>
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