[英]mysqli_query returns false with error code 0, but query succeeds
[英]Mysqli_query returns false and Mysqli_error returns NULL?
我一直在创建预订系统并创建约会,但是我的SQL语句不起作用。 我一直在寻找解决方案,但无济于事。
下面列出的是我的PHP代码。 我的第一个SQL语句可以正常工作并返回正确的ClientID,但是,第二个SQL语句并未将其全部插入数据库中。 我对结果做了var_dumps,返回bool(false),对结果做了mysqli_error,返回null。 我的错误消息最后仅显示echo'd消息,也没有mysqli_error或错误号。
(注意:为了保护数据,某些值已更改/删除)
<?php
session_start();
if(! $_SESSION['Username']) {
header("location:Index.php");
}
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "";
$tablename = "appointmentinformation";
$tablenamed = "clientinformation";
$connection = mysqli_connect("$servername", "$username", "$password", "$dbname") or die("Could not connect to the database");
$clientusername = $_SESSION['Username'];
$sql = "SELECT ClientID FROM $tablenamed WHERE Username = '$clientusername' LIMIT 1";
$results = mysqli_query($connection, $sql);
if (! $results) {
echo ("Could not select the data : " . mysql_error());
} else {
$datarows = mysqli_fetch_row($results);
$clientid = $datarows[0];
}
$date = $_POST["Date"];
$month = $_POST["Month"];
$year = $_POST["Year"];
$time = $_POST["Time"];
$length = $_POST["Length"];
$date = stripslashes($date);
$month = stripslashes($month);
$year = stripslashes($year);
$time = stripslashes($time);
$length = stripslashes($length);
$date = mysqli_real_escape_string($date);
$month = mysqli_real_escape_string($month);
$year = mysqli_real_escape_string($year);
$time = mysqli_real_escape_string($time);
$length = mysqli_real_escape_string($length);
$query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$clientid', '$length', '$date', '$month', '$year', '$time', '$price')";
$result = mysqli_query($connection, $query);
if ($result) {
header("Location:UserCP.php");
} else {
echo ("Could not insert data : " . mysqli_error($result) . " " . mysqli_errno());
}
?>
$query = "INSERT INTO appointmentinformation (ClientID, Length, Date, Month, Year, Time, Price) VALUES ('$clientid', '$length', '$date', '$month', '$year', '$time', '$price')";
$result = mysqli_query($connection, $query);
if ($result) {
header("Location:UserCP.php");
} else {
echo ("Could not insert data : " . mysqli_error($result) . " " . mysqli_errno());
}
mysqli_query返回的结果为空。 这会将您发送到代码的else分支。 然后,您编码了mysqli_error($ result)等于mysqli_error(null)。
我读过的文档说要使用与查询的“链接”初始化变量。 您这样做的原因是:
$connection = mysqli_connect("$servername", "$username", "$password", "$dbname") or die("Could not connect to the database");
现在,您要将其编码为mysqli_error($ connection)和mysqli_errno($ connection)。
另一个建议。 在mysqli_connect语句之后立即添加此代码。
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
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