Parse error: syntax error, unexpected T_STRING on line 6

Question

I keep getting a

Parse error: syntax error, unexpected T_STRING in /newref-exec.php on line 6

I have read loads of different solutions to this problem but non seem to work

<?php
// connect to database
 require($DOCUMENT_ROOT . "connect.php");
// check for blank entries
if ($_POST[doi2] == "NULL"){
    echo "No Data to add";
    }
    else
    {
    $sql="INSERT INTO ref (uid, date, doi, title, year, journal)
    VALUES 
    ('1','CURDATE ()','$_POST[doi2]','$_POST[title2]','$_POST[year2]','$_POST[journal2]')";
    if (!mysqli_query($link,$sql,$con))
        {
        die('The Reference could not be added, beacuse of SQL Error:' . mysql_error());
        }
        echo "New Reference Added";
    }
?>

What is $DOCUMENT_ROOT, I hope you mean $_SERVER['DOCUMENT_ROOT'] (register_globals should never be on). — gen_Eric
– gen_Eric, Commented Oct 23, 2012 at 16:06

Teena Thomas · Accepted Answer · 2012-10-23 16:01:33Z

2

This line, the key needs to have quotes,

if ($_POST['doi2'] == "NULL"){

Also, if you want to check for empty entries, you need to probably do this,

if ($_POST['doi2'] == ""){ //checking "NULL", checks for a string 'NULL'

answered Oct 23, 2012 at 16:01

Teena Thomas

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2 Comments

gen_Eric Over a year ago

It doesn't need to, it should. This will throw a warning, not an error.

Abhishek Saha Over a year ago

There are a lot of other places where quotes is required. And quotes is not required when comparing with null.

Jocelyn · Accepted Answer · 2012-10-24 16:39:36Z

0

Rewrite the code in the below way.

<?php
// connect to database
 require($DOCUMENT_ROOT . "connect.php");
// check for blank entries
if ($_POST['doi2'] == null){
    echo "No Data to add";
} else {
        $doi2 = $_POST['doi2'];
        $title2 = $_POST['title2'];
        $year2 = $_POST['year2'];
        $journal2 = $_POST['journal2'];

        $sql="INSERT INTO ref(uid, date, doi, title, year, journal) VALUES('1', 'CURDATE()', '$doi2', '$title2', '$year2', '$journal2')";
        if (!mysqli_query($link, $sql, $con))
        {
            die('The Reference could not be added, beacuse of SQL Error:' . mysql_error());
        }
        echo "New Reference Added";
}
?>

edited Oct 24, 2012 at 16:39

Jocelyn

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answered Oct 23, 2012 at 16:06

Abhishek Saha

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4 Comments

Abhishek Saha Over a year ago

You have a lot of quotes problem. I changed the $_POST from your query and gave variables which i defined before your query.

Abhishek Saha Over a year ago

Also you dont need quotes when comparing with null.

Matt S Over a year ago

!empty() would be preferable to == null.

Abhishek Saha Over a year ago

In many cases, empty() cannot be used with string variables for the simple fact that empty() returns true if your variable is set the the string value '0'. If the string character '0' is a possibly valid value for your string variable — if all non-zero-length strings are valid — you cannot use empty(). This is even more dangerous if you are using empty() to check if a string variable is defined. Also worth mentioning, empty() returns false for a string value that is nothing but a single space (or any number of spaces really).

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