我的SQL查询有什么问题？

Question

I have a bunch of database calls in my code.

One of them is like this:

mysql_query("INSERT INTO coupons (id, retailerName, description, savingsDetails, pictureURL, rating, qrPicture, zipCode, dateAdded, dateExp) VALUES ('$newID', '$retailerName', '$description', '$savingsDetails', '$pictureURL', '0', '$qrPicture', '$zipCode', '$dateAdded', '$dateExp')");

All of the other database calls works perfectly, but this one simply doesn't write any data onto the table. What's wrong with it?

Answer 1

Check the return from mysql query.

$result = mysql_query($query);
if ($result === false) {
    // Prints the query and the mysql error to help you find the problem.
    die($query.'<br/>'.mysql_error());
}

If there is something wrong with the query it will tell you what it was.

mysql_query() returns true on success and false on error for queries that don't return something.

http://php.net/manual/en/function.mysql-query.php

When using mysql_query() in production, you should check the return value, and exit gracefully on error, and preferably log it so that it can help you fix any issues.

Also when performing a query that uses user input ensure you use mysql_real_escape_string() to avoid SQL injection, and will protect input that contains a single quote, which would break your current query.

http://php.net/manual/en/function.mysql-real-escape-string.php

Answer 2

Check the return value from the query to find out what is happening -

$result = mysql_query("INSERT INTO coupons (id, retailerName, description, 
                savingsDetails, pictureURL, rating, qrPicture, zipCode, dateAdded, 
                dateExp) VALUES ('$newID', '$retailerName', '$description', 
                '$savingsDetails', '$pictureURL', '0', '$qrPicture', '$zipCode', 
                 '$dateAdded', '$dateExp')");

if($result)
{
    // Your query succeeded
}
else
{
    // Your query failed due to some error
    die( mysql_error() );
}

Answer 3

You can just write this as mysql_query("INSERT INTO coupons (id, retailerName, description, savingsDetails, pictureURL, rating, qrPicture, zipCode, dateAdded, dateExp) VALUES ('$newID', '$retailerName', '$description', '$savingsDetails', '$pictureURL', '0', '$qrPicture', '$zipCode', '$dateAdded', '$dateExp')") or die(mysql_error());

If you don't get an error it means it succeeded. If you get an error, you'll an error straight from mySQL (so it should be pretty detailed).

Answer 4

Another way to locate your problem: logging your sql query, to see what the query looks like at run-time.

$str = "INSERT INTO coupons ...";

echo $str;

Then try that query to your database directly, you will get the things that doesn't right.