[英]mysqli_insert_id returning 0
I have to use a reference to a record in another table (FK). 我必须使用对另一个表(FK)中的记录的引用。 The record in the other table may or not exist, so I can either use its ID or insert a new one and then use its ID.
另一个表中的记录可能存在或不存在,因此我可以使用其ID或插入新的ID,然后使用其ID。
I have the following code: 我有以下代码:
$con=mysqli_connect("localhost","root","test","db_site");
// get existing levels
$levelIdSQL = "SELECT idlevel from levels where levelstring = $level";
$levels = mysqli_query($con, $levelIdSQL);
$levelID = -1;
if ($levels['num_rows'] > 0) {
echo "<br><br>some results<br><br>";
// we already have a level, use its id when updating
$row = $levels->fetch_assoc();
$levelID = $row['idlevel'];
} else {
// we must insert this string in the levels table, then use its id
echo "<br>running query: " . "INSERT INTO `db_site`.`levels` ('levelstring') values ('$level');";
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
echo "<br><br>connected OK<br><br>";
}
$rez = mysqli_query($con, "INSERT INTO `db_site`.`levels` ('levelstring') values ('$level');");
$levelID = mysqli_insert_id($con);
}
echo "<br><br>LEVEL ID: " . $levelID;
My levels table has an autoinc "idlevel" field and running the same SQL with MySQL Workbench/command line interface inserts the record just fine. 我的关卡表中有一个autoinc“ idlevel”字段,并且使用MySQL Workbench /命令行界面运行相同的SQL即可插入记录。 However, mysqli_insert_id returns 0 and no records are inserted.
但是, mysqli_insert_id返回0,并且不插入任何记录。
What am I doing wrong? 我究竟做错了什么?
Edit: after jterry's suggestion, I called die() and it returned: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''levelstring')". 编辑:在jterry的建议之后,我调用die()并返回:“您的SQL语法有错误;请检查与您的MySQL服务器版本相对应的手册,以找到在'levelstring')附近使用的正确语法”。 So I changed this:
所以我改变了这个:
$rez = mysqli_query($con, "INSERT INTO `db_site`.`levels` ('levelstring') values ('$level');");
into this: 到这个:
$rez = mysqli_query($con, "INSERT INTO `db_site`.`levels` (levelstring) values ('$level');");
Everything is ok now. 现在一切都很好。
try this , you should use backticks around column name not quotes. 尝试此操作,您应该在列名(而不是引号)周围使用反引号。
$rez = mysqli_query($con, "INSERT INTO `db_site`.`levels` ( `levelstring`)
VALUES ('$level') ");
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