syntax error, unexpected '}'

rstevens

This should be simple, but I can't figure it out.

I want to check if $Corn has any data in the field...

$Corn = $row['Corn'];

Most of the time it does and everything works ok.

But if it does not have any data, I need to change the variable $Corn from nothing to the word "Undisclosed".

I check it this way...

If (!$Corn) {
$Corn = "Undisclosed"
}
else {
$Corn = $english_format_number = number_format($Corn)
};

But I keep getting syntax error, unexpected '}' on this line.

Any suggestions?

Ashley_Sheridan

You've mis-placed your semicolon, it should be after the ) on the second from last line you posted, not after the }

The semi-colon marks the end of a statement, and the braces ({) enclose statements, so you're attempting to close a statement block before ending the statement.

rstevens

Thanks for the quick response! I tried that originally, but I still get the same error message.

rstevens

It says I get the error on line 26...

14 $Corn = $row['Corn'];
15 $Soybeans = $row['Soybeans'];
16 $Wheat = $row['Wheat'];
17 $Total = $row['Total'];
18 $DairyCows = $row['DairyCows'];
19 $CattleOnFeed = $row['CattleOnFeed'];
20 $HogAndPig = $row['HogAndPig'];
21 $Broiler = $row['Broiler'];
22 $Turkey = $row['Turkey'];
23 $Layer = $row['Layer'];
24 If (!$Corn) {
25 $Corn = "Undisclosed"
26 }
27 else {
28 $Corn = $english_format_number = number_format($Corn);
29 }

Ashley_Sheridan

You need to put a semicolon at the end of the first statement in the if too. Hint, it's the line right before 26 😉

rstevens

Thank you very much!

NogDog

The lesson here is that the line number in a PHP syntax error message tells you where the parser was when it threw up its hands and gave up -- not necessarily where the actual error was. 😉