cant add POST content to mysql database

Question

<?php

require "config.php";

/*
CREATE TABLE  `addnews` (
 `id` INT NOT NULL AUTO_INCREMENT PRIMARY KEY ,
 `auther` VARCHAR( 255 ) NOT NULL ,
 `title` VARCHAR( 255 ) NOT NULL ,
 `content` LONGTEXT NOT NULL
) ENGINE = MYISAM ;
*/

$a = $_POST['author'];
$t = $_POST['title'];
$c = $_POST['content'];

if(isset($_POST["add"]) and $_POST["add"] == "news"){
    $insert = mysql_query('INSERT INTO addnews 
    (author,title,content)
    VALUES
    ("$a","$t","$c")') or die("error");
    if (isset($insert )){
        echo "<h3>Done</h3>";
    }
};

echo "
<form action='".$_SERVER['PHP_SELF']."' method='post'>
Author : <input type='text' name='author' /><br>
Title : <input type='text' name='title' /><br>
Content : <textarea name='content'></textarea>
<input type='submit' value='Add news' />
<input type='hidden' name='add' value='news' />
</form>
";


mysql_close($connectdb);
?>

i am getting error from this statment i think

if(isset($_POST["add"]) and $_POST["add"] == "news"){
    $insert = mysql_query('INSERT INTO addnews 
    (author,title,content)
    VALUES
    ("$a","$t","$c")') or die("error happend while trying to add information to database");
    if (isset($insert )){
        echo "<h3>Done</h3>";
    }
};

output is : error happend while trying to add information to database

and no problem with config.php file (the file that connect to database) i am using phpmyadmin

Read this, please: http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?lq=1 — vhu, Jun 30 '15 at 06:58
i dont care about security for now this is my first sql code i am just learning — Asad Asa, Jun 30 '15 at 07:05
Especially in that case I'd suggest to learn it the _right_ way from the beginning. — vhu, Jun 30 '15 at 07:22

score 0 · Answer 1 · answered Jun 30 '15 at 06:54

Use && instead of the actual word and:

if(isset($_POST["add"]) && $_POST["add"] == "news"){
    $insert = mysql_query("INSERT INTO addnews 
    (author,title,content)
    VALUES
    ('$a','$t','$c')") or die("error happend while trying to add information to database");
    if (isset($insert )){
        echo "<h3>Done</h3>";
    }
};

score 0 · Answer 2 · answered Jun 30 '15 at 06:56

0

Here you go try this one

if(isset($_POST["add"]) and $_POST["add"] == "news"){
    $insert = mysql_query('INSERT INTO addnews 
    (author,title,content)
    VALUES
    ("'. $a .'","'. $t .'","'. $c .'")') or die("error happend while trying to add information to database");
    if (isset($insert )){
        echo "<h3>Done</h3>";
    }
};

used "'. $a .'" instead "$a".

answered Jun 30 '15 at 06:56

Ankit Pise

1,243
11
30

still the sam problem :( – Asad Asa Jun 30 '15 at 07:02
you sure you're receiving post values correctly, try `var_dump($_POST)` – Ankit Pise Jun 30 '15 at 07:18

score 0 · Answer 3 · answered Jun 30 '15 at 07:06

0

I think the query statement is wrong, Double quotes inside the single quotes is not valid in php. So you will change the quotes in query like below code,

$insert = mysql_query("INSERT INTO addnews 
    (author,title,content)
    VALUES
    ('$a','$t','$c')") or die("error");

try this..:-)

answered Jun 30 '15 at 07:06

sathish kumar

116
1
8

$insert = mysqli_query("INSERT INTO addnews (author,title,content) VALUES ('".$a."','".$t."','".$c."')") or die("error"); – sathish kumar Jun 30 '15 at 07:15
try mysqli_query() instead of mysql_query() – sathish kumar Jun 30 '15 at 07:16

score 0 · Answer 4 · answered Jun 30 '15 at 07:24

Please do the correction in your code like as follow:

$insert = mysql_query("INSERT INTO addnews 
    (author,title,content)
    VALUES
    ('$a','$t','$c')") or die(mysql_error($link));//Where $link mysql resource object

You will get the answer why Mysql not inserting your data.

score 0 · Accepted Answer · answered Jun 30 '15 at 07:25

strings in sql are surrounded by ' (single quote) , not by " (double quote)
strings in php will act two ways
1. those in ' (single quote) will write literally as tyou typed them ($a stays $a - not $a value)
2. those in " (double quote) will interpret values inside - so $a will be substituted with $a's value
when failing DB operation - it is usually useful to see what was wrong - use mysql_error for that

cant add POST content to mysql database

output is : error happend while trying to add information to database

5 Answers5