Error Unexpected T_VARIABLE

Question

Ok I've been messing with this code for quite a while now and it keeps saying I have an unexpected T_VARIABLE on line 54. Anyone know what it is and how to fix?

// Write the key and activation time to the database as a new row
**LINE 54**   $registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES("$key","$time")") or die(mysql_error());

May be error here...write this way->Values('".$key.",'".$value."') $registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES("$key","$time")") or die(mysql_error()); — Butani Vijay, Jul 20 '13 at 10:26
Now you've had this question answered, forget about MySQL and start learning MySQLi or PDO with prepared statements: you're clearly just starting to learn, so learn the better methods of querying databases, rather than bad methods that you'll have to unlearn later — Mark Baker, Jul 20 '13 at 10:26
One can't stress @MarkBaker's point enough. Please stop concatenating strings to make queries and start using prepared statements already. — Pranav Negandhi, Jul 20 '13 at 10:28

score 1 · Answer 1 · edited May 25 '19 at 14:21

You need to use single quotes instead of double quotes:

$registerid = mysql_query("
INSERT INTO downloadkey (uniqueid,timestamp) 
VALUES('$key','$time')") or die(mysql_error());

The other solution is to concatenate your variables, like so:

$registerid = mysql_query("
INSERT INTO downloadkey (uniqueid,timestamp) 
VALUES(" . $key . "," . $time . ")") or die(mysql_error());

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

score 0 · Answer 2 · answered Jul 20 '13 at 10:25

0

There was Syntax error. The code was broken by the use of double quotes near VALUES("$key","$time")

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES('$key','$time')") or die(mysql_error());

answered Jul 20 '13 at 10:25

curious_coder

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It happens. Thats why we are here to help. – curious_coder Jul 20 '13 at 10:32

score 0 · Answer 3 · answered Jul 20 '13 at 10:25

your insert query looks like it has been broken by the " php is smart enough can read its variable inside the " quotes then no need to wrap you variables in the " when they are already opened

 $registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp)
 VALUES($key,$time)") or die(mysql_error());

score 0 · Answer 4 · answered Jul 20 '13 at 10:25

0

Correct line 54 is

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES(\"$key\",\"$time\")") or die(mysql_error());

answered Jul 20 '13 at 10:25

flavien.crochard

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score 0 · Answer 5 · answered Jul 20 '13 at 10:25

0

**LINE 54**   $registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES(\"$key\",\"$time\")") or die(mysql_error());

Escape your quotes.

answered Jul 20 '13 at 10:25

Pranav Negandhi

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score 0 · Answer 6 · answered Jul 20 '13 at 10:26

This line is the culprit. If you break off a string to insert variables you need to concatenate them

Solution A:

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES('".$key."','".$time."')") or die(mysql_error());

Solution B

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES('$key','$time')") or die(mysql_error());

And please don't use deprecated functions like mysql_*

score 0 · Answer 7 · answered Jul 20 '13 at 10:26

0

maybe try:

$registerid = mysql_query("INSERT INTO downloadkey (`uniqueid`,`timestamp`) VALUES('$key','$time')") or die(mysql_error());

answered Jul 20 '13 at 10:26

pulsar

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score 0 · Answer 8 · answered Jul 20 '13 at 10:26

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES("$key","$time")") or die(mysql_error());

Replace with:

$registerid = mysql_query("INSERT INTO downloadkey (uniqueid,timestamp) VALUES('" . $key . "','" . $time . "')") or die(mysql_error());

Error Unexpected T_VARIABLE

8 Answers8