[Dr.Jamescook]

Hello,
I am having trouble finding the error in my SQL syntax. Here is what it says:

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 'FROM nuke_bible WHERE Book = 'Deuteronomy' AND Chapter = '20''

Here is my code:

PHP Code:

mysql_query("SELECT Testament, Book, Chapter, Texta, Textb, Textc, $ID $TMID $VerseID FROM nuke_bible WHERE Book = '$book' AND Chapter = '$chapter'") or die(mysql_error()); 

I have tried everything that I know is possible to try. Please help. Thanks.

[NeverMind]

I am suspecting the part $ID $TMID $VerseID!
print the query and show us the result.

[FiRe]

$ID $TMID $VerseID

are these seperate variables? in which case it shud be...

$ID, $TMID, $VerseID

[Dr.Jamescook]

Here is the whole code so far. It was the only way I could think to do it. My point was if they selected ID checkbox then pull the column ID out. If not it won't print anything there so the sql_query does know that there was something there.

PHP Code:

 $VerseID = $_POST['VerseID'];
 $fontsize = $_POST['fontsize'];
 $fonttype = $_POST['fonttype'];
 $submit = $_POST['Submit'];
 
 echo $book . "<br>";        // For Debugging.
 echo $chapter . "<br>";        // For Debugging.
 echo $ID . "<br>";            // For Debugging.
 echo $TMID . "<br>";        // For Debugging.
 echo $VerseID . "<br>";        // For Debugging.
 echo $fontsize . "<br>";    // For Debugging.
 echo $fonttype . "<br>";    // For Debugging.
 echo $submit . "<br>";        // For Debugging.
 
 //**************************//
 //   Start Error Checking    //
 //**************************//
 
 if($book == "Please select:") {
     echo ("Please select a book.");
 }
 
 if($chapter == "") {
     echo ("Please type a chapter.");
 }
 //**********************//
 //  End Error Checking    //
 //**********************//
 
 
 //******************************************//
 //  Start Defining Variables for SQL query    //
 //******************************************//
 
 if(isset($ID)) {
     $ID = "ID,";
 } else {
     $ID = '';
 }
 echo "<p>" . "ID equals: " . $ID . "<br>";    // For Debugging.
 
 if(isset($TMID)) {
     $TMID = "TMID,";
 } else {
     $TMID = "";
 }
 echo "TMID equals: " . $TMID . "<br>";        // For Debugging.
 
 if(isset($VerseID)) {
     $VerseID = "VerseID,";
 } else {
     $VerseID = "";
 }
 echo "VerseID equals: " . $VerseID . "<br>";// For Debugging.
 //******************************************//
 //   End Defining Variables for SQL query    //
 //******************************************//
 
 
 //******************************************//
 // Start Defining Variables for Text Styles //
 //******************************************//
 
 if($fonttype == "Arial") {
     $fonttype = "Arial, Helvetica, san-serif";
 } elseif($fonttype == "Times New Romans"){
     $fonttype = "Times New Roman, Times, serif";
 } elseif($fonttype == "Courier") {
     $fonttype = "Courier New, Courier, mono";
 } elseif($fonttype == "Georgia") {
     $fonttype = "Georgia, Times New Roman, Times";
 } elseif($fonttype == "Verada") {
     $fonttype = "Vernada, Arial, Helvetica, sans-serif";
 } elseif($fonttype == "Geneva") {
     $fonttype = "Geneva, Arial, Helvetica, sans-serif";
 }
 echo "Fonttype equals: " . $fonttype . "<br>";     // For Debugging.
 //******************************************//
 //  End Defining Variables for Text Styles     //
 //******************************************//
 
 $user = "root";
 $pass = "";
 mysql_connect('localhost', $user, $pass) or die(mysql_error());
 mysql_query("SELECT Testament, Book, Chapter, Texta, Textb, Textc, $ID $TMID $VerseID FROM nuke_bible WHERE Book = '$book' AND Chapter = '$chapter'") or die(mysql_error()); 

I will be back with some more info in a bit.

[Dr.Jamescook]

Ok, done some more checking. If I remove those variables there is no diffrence. I still can't seem to understand why it is doing this.

[Fallen Angel]

You forgot to select your database.

Just add this after mysql_connect:

PHP Code:

mysql_select_db('my_database'); 

Also put commas after your variables : $ID, $TMID, $VerseID

[Dr.Jamescook]

That doesn't fix that part though. The one I mentioned ^^^ there.

[KaraJ]

You need to put quotes around your variable names, like this:

mysql_query("SELECT Testament, Book, Chapter, Texta, Textb, Textc, `$ID`, `$TMID`, `$VerseID` FROM nuke_bible WHERE Book = '$book' AND Chapter = '$chapter'") or die(mysql_error());

This gave me lots of trouble, but I finally found that you need to use the single quote on the tilde key (upper left on keyboard) when specifying tables to select from and the single quote on the double quote key (right hand side of keyboard) when specifying what to set the tables equal to. Hope this helps!

Kara