mysql php displays results but not inserting into database

Question

I stripped down query for only one insert

<?php
session_start();
include 'cstring.php';
$title="";
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else {
    $title=$_POST['title']; 
    $query=mysqli_query($con,"insert into blogpages(blogpagetitle) values('".$title."')");
    if($query){
        $bloga="sucessfully added a new blog";
        echo $bloga; 
     }
     else {
        echo mysqli_error($con); // if using mysqli do not use mysql in between
     }
}
mysqli_close($con);
?>

is there something wong in this code that it doesnt insert into mysql

table structure

1.bpid int(50)--------------null-no default-none autoincrement

2.blogpagetitle------------varchar(255) utf16_general_ci

3.datemade-------------timestamp current time stamp

4.blogpagedescription---------text utf16_general_ci

5.blogbody----------------longtext utf16_general_ci

6.blogpageextended------------ text utf16_general_ci

To display errors correctly you need to have: print mysqli_error($con); — Mm-Art-In
– Mm-Art-In, Commented Nov 10, 2014 at 11:14
Also I really hope this code is much simplified for SO. NEVER take a post value and insert it straight into a DB, always use something like: $title = mysqli_real_escape_string($_POST['title']); — Mm-Art-In
– Mm-Art-In, Commented Nov 10, 2014 at 11:16
To print the error you should use mysqli_error($con) and not mysql_error(). — Davide Pastore
– Davide Pastore, Commented Nov 10, 2014 at 11:32

MixedVeg · Accepted Answer · 2014-11-10 11:36:52Z

1

TIP

Sanitize variables, Use mysqli_real_escape_string()
When you are not able to debug your code, echo every possible stuff and die the rest code. For example here, echo if there is error in DB connection, echo the query to see if it is correct, echo the result of query execution, echo if there is some error!

edited Nov 10, 2014 at 11:36

answered Nov 10, 2014 at 11:30

MixedVeg

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2 Comments

Davide Pastore Over a year ago

No no. echo mysql_error();} is closing else { $title=$_POST['title']; Check it yourself.

MixedVeg Over a year ago

indentation! God! Let me keep the tips and delete the answer!

Simon Fredriksson · Accepted Answer · 2014-11-10 11:32:11Z

0

You should be using echo mysqli_error($con) to get the error message rather than mysql_error().

answered Nov 10, 2014 at 11:32

Simon Fredriksson

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