Parse error: syntax error, unexpected T_VARIABLE in

[benson.tairuakinsola]

I am trying to set up a json object.array to retrieve the data from this mysql query. the data required is posted from java and that seems to work fine, but I get parsing errors. I tried testing the php file and I get the "Parse error: syntax error, unexpected T_VARIABLE in C:\wamp\www\GetVerse.php on line 8" . previously I had a ';' at the end of the sql query and it said unexpected ';' at the end of line 7. now it i get the error message for line 8: $row = mysql_fetch_array($result); . What seems to be the problem?

<?php
mysql_connect("localhost","root","password");
mysql_select_db("hftwmvirtualdb");
$Booknum = mysql_real_escape_string($_POST['Booknum']); 
$Chapternum = mysql_real_escape_string($_POST['Chapternum']); 
$Versenum = mysql_real_escape_string($_POST['Versenum']); 
$result = mysql_query("SELECT VERSETEXT FROM booktable WHERE BIBLEID = '".$Booknum."' AND CHAPTERID = '".$Chapternum."' AND VERSENO = '".$Versenum."' LIMIT 0,1"
$row = mysql_fetch_array($result);
echo json_encode($row); 
mysql_close();  
?>

[blocblue]

You missed a semi-colon at the end of line 7.

[benson.tairuakinsola]

Is it possible to check whether the php script is correct without the running of the java code to post the required data?

[blocblue]

You're missing a closing bracket too. );

[benson.tairuakinsola]

ahh, can't believe i missed that. I will check if it runs now.

[pritaeas]

Is it possible to check whether the php script is correct

If you use a smart IDE, then you can prevent such typos easily. See this thread.

[benson.tairuakinsola]

I corrected that error & After checking the script through a browswer I'm now receiving the following warning messages:

• 
  
  1. Notice: Undefined index: Booknum in C:\wamp\www\GetVerse.php on line 4
• 
  
  1. Notice: Undefined index: Chapternum in C:\wamp\www\GetVerse.php on line 5
• 
  
  1. Notice: Undefined index: Versenum in C:\wamp\www\GetVerse.php on line 6
• 
  
  1. Notice: Undefined variable: sql in C:\wamp\www\GetVerse.php on line 8
• 
  
  1. Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\wamp\www\GetVerse.php on line 8
• 
  
  1. Notice: Undefined variable: output in C:\wamp\www\GetVerse.php on line 9

I assume, 1-3 are because I did not submit the post data but does is the same reason for messages 4-6?

[blocblue]

1 - 3, you're correct.

5, will be because your query's incorrect due to the missing $_POST parameter values.

4 and 6 do not seem to appear to correspond with the code you have posted, so it's not possible to tell. I cannot see variables called $sql or $output.

[benson.tairuakinsola]

I have corrected a table name in the sql and updated the PHP code. I have previously created a similar php script that gets all data in a table & encodes in Json; with line 8-11 the same. Why is is causing so trivial when i'm trying to mysql query to select a cell. The new PHP Code as stated:

<?php
mysql_connect("localhost","root","password");
mysql_select_db("hftwmvirtualdb");
$Booknum = mysql_real_escape_string($_POST['Booknum']); 
$Chapternum = mysql_real_escape_string($_POST['Chapternum']); 
$Versenum = mysql_real_escape_string($_POST['Versenum']); 
$sql = mysql_query("SELECT 'VERSETEXT' FROM 'booktable' WHERE 'BOOKID' = $Booknum AND 'CHAPTERID' = $Chapternum AND 'VERSENO' = $Versenum");
while($row=mysql_fetch_assoc($sql)) $output[]=$row;
  print(json_encode($output));
  mysql_close();
?>

I thought output was defined in line 8: "while($row=mysql_fetch_assoc($sql)) $output[]=$row;" so that it can be encoded by Json array/object. If this code is incorrect, How do I get the result of the query for parsing?

[benson.tairuakinsola]

ok so after adding echo mysql_error(); before the while statement. I recieved:
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 ''booktable' WHERE 'BOOKID' = '' AND 'CHAPTERID' = AND 'VERSEID' =' at line 1
Which means the $sql didn't read the post data. I thought that was the correct syntax

[pritaeas]

$sql = mysql_query("SELECT 'VERSETEXT' FROM 'booktable' WHERE 'BOOKID' = $Booknum AND 'CHAPTERID' = $Chapternum AND 'VERSENO' = $Versenum");

Should be:

$sql = mysql_query("SELECT `VERSETEXT` FROM `booktable` WHERE `BOOKID` = $Booknum AND `CHAPTERID` = $Chapternum AND `VERSENO` = $Versenum");

In short, replace the single quotes with backticks. Use single quotes around values, backticks around tables and columns.

Since the POST is empty, show us your form or ajax call too.

[Veera_2]

            [02-Aug-2018 04:37:37 Etc/GMT] PHP Parse error:  syntax error, unexpected '$merchant_data' (T_VARIABLE) in /home/turiyavilla/public_html/pay/ccavRequestHandler.php on line 11
[23-Dec-2019 06:19:03 Etc/GMT] PHP Parse error:  syntax error, unexpected '/', expecting end of file in /home/turiyv6z/public_html/pay/ccavRequestHandler.php on line 11
[23-Dec-2019 06:22:05 Etc/GMT] PHP Parse error:  syntax error, unexpected '/', expecting end of file in /home/turiyv6z/public_html/pay/ccavRequestHandler.php on line 11

                <html>
                <head>
                <title> Non-Seamless-kit</title>
                </head>
                <body>
                <center>

                <?php include('Crypto.php')?>
                <?php 
                    error_reporting(0);
                    $merchant_data='';
                    $working_key=' ';//Shared by CCAVENUES
                    $access_code=' ';//Shared by CCAVENUES

                    foreach ($_POST as $key => $value){
                        $merchant_data.=$key.'='.$value.'&';
                    }

                    $encrypted_data=encrypt($merchant_data,$working_key); // Method for encrypting the data.

                ?>
                <form method="post" name="redirect" action="https://secure.ccavenue.com/transaction/transaction.do?command=initiateTransaction"> 
                <?php
                echo "<input type=hidden name=encRequest value=$encrypted_data>";
                echo "<input type=hidden name=access_code value=$access_code>";
                ?>
                </form>
                </center>
                <script language='javascript'>document.redirect.submit();</script>
                </body>
                </html>