Inserting data into mysql table using PHP [duplicate]

Question

$dbc=mysql_connect('127.0.0.1', 'root', '1234','aliendatabase')
or die('Failed!');

$query = "INSERT INTO alien_abduction(first_name, last_name, when_it_happened, how_long, " .
"how_many, alien_description, what_they_did, fang_spotted, other, email) " .
"VALUES ('$first_name', '$last_name', '$when_it_happened', 
'$how_long', '$how_many', " .
"'$alien_description', '$what_they_did', '$fang_spotted', '$other', 
'$email')"; 

$result=mysql_query($query)
or die("Failed to upload!!!!");

mysql_close($dbc);

This code is unable to execute $result line (so outputs Failed to upload!!!!) but it is able to establish connection. I have cross-checked the table column name and variables and it seems fine.

MySQL version 5.7

WARNING: If you're just learning PHP, please, do not use the mysql_query interface. It’s so awful and dangerous that it was removed in PHP 7. A replacement like PDO is not hard to learn and a guide like PHP The Right Way explains best practices. Your user data is not properly escaped and there are SQL injection bugs and can be exploited. — tadman
– tadman, Commented Dec 21, 2017 at 19:22
If only there were a way to get more info about the nature of the error. — Bill Karwin
– Bill Karwin, Commented Dec 21, 2017 at 19:25
Okay, I will learn it but can I know why this code doesn't run? I was referring to a book called Head First PHP & MySQL. — Aman Sharma
– Aman Sharma, Commented Dec 21, 2017 at 19:30
A book? Wait, what is a BOOK?!?! ... :D Switch all your mysql_* to mysqli_* at the very least. — IncredibleHat
– IncredibleHat, Commented Dec 21, 2017 at 19:39
@IncredibleHat It doesn't work. I have tried it. I didn't even connect to the server. — Aman Sharma
– Aman Sharma, Commented Dec 21, 2017 at 19:41

somanalytics · Accepted Answer · 2017-12-21 19:34:26Z

0

Please check the Datatypes of the columns and do the following to find the error:

$result=mysql_query($query) or die(mysql_error($dbc));

answered Dec 21, 2017 at 19:34

somanalytics

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6 Comments

Aman Sharma Over a year ago

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 ' , , , , , , no, , )' at line 1 Which line 1 is it referring to?

somanalytics Over a year ago

Aman, can you please do an echo of the SQL and paste it here

Aman Sharma Over a year ago

echo 'Thanks for submitting the form. '; echo 'You were abducted ' . $when_it_happened; echo ' and were gone for ' . $how_long . ' '; echo 'Number of aliens: '.$how_many.' '; echo 'Describe them: ' . $alien_description . ' '; echo 'The aliens did this: '.$what_they_did." "; echo 'Was Fang there? ' . $fang_spotted . ' '; (Error is in this line as I figured out.) echo 'Other Comments: '.$other.' '; echo 'Your email address is ' . $email;

Aman Sharma Over a year ago

I hope it is readable or I can upload an image.

somanalytics Over a year ago

try: $query = "INSERT INTO alien_abduction(first_name, last_name, when_it_happened, how_long, " . "how_many, alien_description, what_they_did, fang_spotted, other, email) " . "VALUES ('".stripslashes($first_name)."', '".stripslashes($last_name)."', '".stripslashes($when_it_happened)."', '".stripslashes($how_long)."', '".stripslashes($how_many)."', '".stripslashes($alien_description)."', '".stripslashes($what_they_did)."', '".stripslashes($fang_spotted)."', '".stripslashes($other)."', '".stripslashes($email)."')";

|

SamHecquet · Accepted Answer · 2017-12-21 20:18:34Z

-2

The problem is with your quoting way. As single quote won't be used to hold variables. So in single quote, they are simply strings, not the variables.

If you want that $variable replaced with value then rewrite the whole query in single quote like below

$query = 'INSERT INTO alien_abduction(first_name, last_name, when_it_happened, how_long, how_many, alien_description, what_they_did, fang_spotted, other, email) '.
 ' VALUES (' ".$first_name." ', ' ".$last_name . " ', ' ". $when_it_happened . " ', ' " .$how_long . " ', ' " . $how_many . " ', " . " " . $alien_description . " ', ' " . $what_they_did . " ', ' " . $fang_spotted . " ', ' " . $other . " ', ' " . $email . " ')' ;

I hope that will solve the issue.

edited Dec 21, 2017 at 20:18

SamHecquet

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answered Dec 21, 2017 at 19:48

Er. Amit Joshi

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5 Comments

Aman Sharma Over a year ago

We can use double quotes everywhere. It isn't a problem as it recognizes the variable too.

Er. Amit Joshi Over a year ago

Problem is single quote not the double quotes

Benjamin Racette Over a year ago

I don't understand why Er. Amit Joshi and I got -1 for actually solving your problem.

GSerg Over a year ago

The string literal saved into $query uses double quotes, so the variables are parsed. The fact that these variables are also surrounded by single quotes inside the double quotes is irrelevant because it is SQL-level quoting, not PHP level. On top of that, even if unparsed variables was the issue, it would simply insert the unparsed variable names into the database without raising an error.

Er. Amit Joshi Over a year ago

Kindly look into this post stackoverflow.com/questions/3446216/… thanks bro for @Benjamin Racette we are not wrong at all. :)

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