Parse error: syntax error, unexpected '"', expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

Question

Parse error: syntax error, unexpected '"', expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\xampp\htdocs\cc_real\3HLR14CMORCGZ0IY8AE7H4JL409RV9\insert.php on line 15.

Showing me this error, and Below is my code that I am trying to use for inserting data into a database.

<?php
echo "Invoked!!!";

$con = mysqli_connect('localhost', 'root', '');
if (!$con)
{
    die('could not connect:'.mysql_error());    
}
mysqli_select_db('job1');


Error: $sql = "INSERT INTO `cc_job2`(cc_Answer_filename,cc_time,cc_workerID)  VALUES
       ('$_post["Answer_filename"]','$_post["track_data"]','$_post["workerID"]')";

$result = mysqli_query($sql)
if ($result){
    echo ("<br> Input data is succeed") 
} else{
    echo("<br> Input data is failed");
}
mysqli_close($con);
?>

The syntax highlighting of your question will give you a clue. — Federkun, Dec 23 '16 at 09:30
you just forgot a closing `;` in some line(s) like `$result = mysqli_query($sql)` and `echo ("
Input data is succeed")` — Beginner, Dec 23 '16 at 09:32
after **$result = mysqli_query($sql) and echo ("
Input data is succeed")** add semicolon `;` — Soni Vimalkumar, Dec 23 '16 at 09:33

score 2 · Answer 1 · edited May 23 '17 at 11:46

There are several issues with this piece of code.

Missing ; at the end of several lines
Missing the $con parameter for most mysqli_ functions
Mixing mysql_ and mysqli_
Using $_post instead of $_POST (capital letters)

See the comments around the code for what's been changed.

<?php
echo "Invoked!!!";

$con = mysqli_connect('localhost', 'root', '');
/*
You can also do this, and drop mysqli_select_db() later in the code
$con = mysqli_connect('localhost', 'root', '', 'job1');
*/

if (!$con) {
    // Cannot mix mysql with mysqli (changed out mysql_error())
    // Also, mysqli has "mysqli_connect_error()" for connecting errors
    die('could not connect: '.mysqli_connect_error());    
}

// This function require the $con parameter
mysqli_select_db($con, 'job1');

// Quotes being messed up - this is your current error
// Concatenate the POST values instead, like this
// Also using $_post instead of $_POST
$sql = "INSERT INTO `cc_job2` (cc_Answer_filename, cc_time, cc_workerID) VALUES ('".$_POST["Answer_filename"]."', '".$_POST["track_data"]."', '".$_POST["workerID"]."')";

// Missing $con as the first parameter and ; at the end
$result = mysqli_query($con, $sql);
if ($result) {
    // Missing ; at the end 
    echo "<br> Input data is succeed";
} else{
    echo "<br> Input data is failed";
    // You should add echo mysqli_error($con); here to troubleshoot queries
}
mysqli_close($con);
?>

Note that the query will fail if any of the POST-values contains singlequotes ', so you should either escape them, or better yet, use prepared statements (see the paragraph below and the link "How can I prevent SQL injection in PHP?" at the bottom.

This code is also vulnerable to SQL injection, and you should use prepared statements with placeholders to guard yourself against this.

See these links

Pravin Vavadiya · Accepted Answer · 2016-12-23T09:34:52.117

Just Add ; end of the line echo ("<br> Input data is succeed") and $result = mysqli_query($sql).

Try to something like this.

<?php
echo "Invoked!!!";

$con = mysqli_connect('localhost', 'root', '');
if (!$con)
{
die('could not connect:'.mysql_error());    
}
mysqli_select_db('job1');


Error: $sql = "INSERT INTO cc_job2(cc_Answer_filename,cc_time,cc_workerID) VALUES ('".$_post['Answer_filename']."','".$_post['track_data']."','".$_post['workerID']."')";

$result = mysqli_query($sql);
if ($result){
echo ("<br> Input data is succeed");

}else{
echo("<br> Input data is failed");
}
mysqli_close($con);
?>

I don't think typeerror is acceptable – Beginner Dec 23 '16 at 09:34 — Beginner, Dec 23 '16 at 09:34

score 0 · Answer 3 · edited Dec 23 '16 at 09:40

0

$sql = "INSERT INTO `cc_job2`(cc_Answer_filename,cc_time,cc_workerID)  VALUES ('".$_post['Answer_filename']."','".$_post['track_data']."','".$_post['workerID']."')";

edited Dec 23 '16 at 09:40

Federkun

36,084
8
78
90

answered Dec 23 '16 at 09:36

Dharati Patel

1
4

try with this syntax – Dharati Patel Dec 23 '16 at 09:37

score 0 · Answer 4 · answered Dec 23 '16 at 09:37

Concate in mysql query:

<?php
echo "Invoked!!!";

$con = mysqli_connect('localhost', 'root', '');
if (!$con)
{
die('could not connect:'.mysql_error());    
}
mysqli_select_db('job1');


Error: $sql = "INSERT INTO cc_job2 (cc_Answer_filename,cc_time,cc_workerID)  VALUES
       ('".$_post["Answer_filename"]."','".$_post["track_data"]."','".$_post["workerID"]."')";

$result = mysqli_query($sql)
if ($result){
echo ("<br> Input data is succeed");

}else{
echo("<br> Input data is failed");
}
mysqli_close($con);
?>

Parse error: syntax error, unexpected '"', expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

4 Answers4