[英]Inserting data from a form to multiple tables in a mysql database using php
[英]MySQL PHP - Inserting form data into tables. Unknown Error
因此,正如标题所述,如果表单通过了一些条件,我将尝试将数据插入2个表中(用户已登录,用户之前尚未对该产品进行投票,所有字段均已填写) 。
当条件刚好时,它将数据完美插入两个表中:
$sql = "SELECT productid FROM votes WHERE username='$username' LIMIT 1";
但是我意识到,如果用户对任何一种产品进行了投票,那么他们的用户名将出现在表格中,并且如果不对该产品进行投票,他们将无法通过该条件。 所以我刚刚添加:
$sql = "SELECT productid FROM votes WHERE username='$username' AND productid='$id' LIMIT 1";
现在,如果我尝试将数据提交到数据库,它将始终返回“将错误插入票表”消息,但不会返回mysql_error(),并且显然不会在票表中插入新行,但是很奇怪会更新产品表。
我只是不知道发生了什么,所以如果有人可以帮助我诊断问题,我将非常感激! 这是代码:
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST'){
if($_POST['slider_surface'] !== "0" && $_POST['slider_edgewear'] !== "0" && $_POST['slider_centering'] !== "0" && $_POST['slider_corners'] !== "0"){
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'root';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$slider_surface = $_POST['slider_surface'];
$slider_edgewear = $_POST['slider_edgewear'];
$slider_centering = $_POST['slider_centering'];
$slider_corners = $_POST['slider_corners'];
$id = preg_replace('#[^a-z0-9]#i', '', $_GET['id']);
session_start();
$username = $_SESSION['username'];
//check if user has already voted
mysql_select_db('products');
$sql = "SELECT productid FROM votes WHERE username='$username' AND productid='$id' LIMIT 1";
$query = mysql_query( $sql, $conn );
$uname_check = mysql_num_rows($query);
if ($username){
if ($uname_check < 1) {
$sql = "INSERT INTO votes ".
"(username,productid,votesurface,voteedgewear,votecentering,votecorners,datetime) ".
"VALUES('$username','$id','$slider_surface','$slider_edgewear','$slider_centering','$slider_corners', now())";
$retval = mysql_query( $sql, $conn );
$id='';
// Make sure the _GET product ID is set, and sanitize it
$id = preg_replace('#[^a-z0-9]#i', '', $_GET['id']);
//Retrieves data from MySQL
$data = mysql_query("SELECT * FROM products WHERE id='$id'") or die(mysql_error());
$product = mysql_fetch_array( $data );
$newvotecount = $product['votecount'] + 1;
$newsum_surface = $product['sumsurface'] + $slider_surface;
$newsum_edgewear = $product['sumedgewear'] + $slider_edgewear;
$newsum_centering = $product['sumcentering'] + $slider_centering;
$newsum_corners = $product['sumcorners'] + $slider_corners;
$sql = "UPDATE products SET votecount='{$newvotecount}', sumsurface='{$newsum_surface}', sumedgewear='{$newsum_edgewear}', sumcentering='{$newsum_centering}', sumcorners='{$newsum_corners}' WHERE id='$id'";
$retval2 = mysql_query( $sql, $conn );
if(! $retval){
die('Error inserting into votes table: ' . mysql_error());
}
else if(! $retval2){
die('Error inserting into products table: ' . mysql_error());
}
$grading_error = 'success';
mysql_close($conn);
} else
$grading_error = 'duplicateuser';
} else
$grading_error = 'nouser';
}
else
$grading_error = 'emptyfields';}
?>
将数据插入表时存在问题,因此INSERT
语句必须存在错误。 正如@ user2910809在以前的注释中所说的, sql
计算结果为:
INSERT INTO votes (username,
productid,
votesurface,
voteedgewear,
votecentering,
votecorners,
datetime)
VALUES ('magmar',
'52',
'7',
'6',
'5',
'5',
now())
这句话在语法上是正确的。 因此,如果有错误,则必须是插入的值。
正如@ user2910809对他的评论指出,有一个UNIQUE
的密钥username
栏,它是什么引发错误。
因此,解决方案是修改表的索引以允许具有相同username
多行。
编辑 :如注释中所建议,解决此问题的SQL是: ALTER TABLE votes DROP INDEX username;
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