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  1. #1
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    Access denied for user 'blue'@'localhost' (using password: YES)

    Hello,

    I know very little about programming, means what ever learn, i learnt it from the you tube, so in order to explain my problem i am writing every thing so when i get the reply i can understand it.

    first i try to download apache, mysql and phpmyadminn but did not succed.

    after that i downloaded XAMPP 1.7.7

    In XAMPP 1.7.7 I gave this information for my security

    MYSQL SECTION: "ROOT" PASSWORD (This line was already there as a heading)

    MYSQL super user : root (it was already there so i cant change)
    password : bluebus
    PhpMyAdmin authentification: http or cokie (i select cokie),
    Safe plain password in text file? ((File: C:\xampp\security\security\mysqlrootpasswd.txt): there was a check box i did not click,

    XAMPP DIRECTORY PROTECTION (.htaccess) (this line was already there as a heading)

    User: bus
    Password : bluebuss,
    Safe plain password in text file? ((File: C:\xampp\security\security\mysqlrootpasswd.txt): there was a check box i did not click,

    after that i went to phpmyadmin

    user : root
    password: redbus

    after this i created a database, DATABASE NAME: CAR,

    After this i added a new user for this database, NEW USER NAME: CAR, and password i gave again "redbus", and i created a table,

    after this i created a new file in notpad++, the coding is

    1. <?php
    2.
    3. define ('DB_NAME', 'car');
    4. define ('DB_USER', 'car');
    5. define ('DB_PASWORD', 'redbus');
    6. define ('DB_HOST', 'localhost');
    7.
    8. $link= mysql_connect (DB_HOST,DB_USER,DB_PASSWORD);

    9. if (!$link) {
    10. die ('could not connect:' . mysql_error () );
    11. }
    12.
    13. $db_selected=mysql_selected_db (DB_NAME, $link);
    14.
    15. if (!$db_selected) {
    16. die ('cant user' . ':' . mysql_error ());
    17. }
    18.
    19. $value = $_post ['input1'];
    20.
    21. $sql= "INSERT INTO demo (input1) VALUES ('$value')";
    22.
    23. if (!mysql_query ($sql)) {
    24. die ('error:' . mysql_error () );
    25. }
    26.
    27. mysql_close ();
    28. ?>


    when i try to run it its give me this error and i dont understand why

    Notice: Use of undefined constant DB_PASSWORD - assumed 'DB_PASSWORD' in C:\xampp\htdocs\PhpProject1\index.php on line 8

    [Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'car'@'localhost' (using password: YES) in C:\xampp\htdocs\PhpProject1\index.php on line 8
    could not connect:Access denied for user 'malakiarif'@'localhost' (using password: YES)



    please help me guys, i try to found out the answer by googling and wasted my whole week but could not find answer, please help me and tell me what i need to correct.

    thanks

  2. #2
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    Rubble's Avatar
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    For a start you are using PASSWORD on line 8 and PASWORD on line 5.

  3. #3
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    thanks i corrected it, but its still showing me the same error,

    or i am doing some other mistake. after your message i check my other spelling mistake as well,

    thanks

    help please

  4. #4
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    and one more thing after correcting my spelling mistake it is only showing one error, so thanks for your help, the error which its showing is

    Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'arif'@'localhost' (using password: YES) in C:\xampp\htdocs\PhpProject1\index.php on line 8
    could not connect:Access denied for user 'arif'@'localhost' (using password: YES)

    thanks

  5. #5
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    i think i was doing some stupid mistake so this time i added a new user name: member, password: redblue and changed the coding accordingly on line 4 and and five now its look like ......
    line 4. define ('DB_USER', 'member');
    line5. define ('DB_PASSWORD', 'redblue');
    but after running it is giving me the error **Fatal Error:** Call to undefined function mysql_selected_db() in C:\xampp\htdocs\PhpProject1\index.php on line 14
    line 14 is
    $db_selected=mysql_selected_db (DB_NAME, $link);
    thanks for help

  6. #6
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    Try changing $db_selected=mysql_selected_db (DB_NAME, $link); to $db_selected=mysql_select_db (DB_NAME, $link);

  7. #7
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    thanks

    Quote Originally Posted by Rubble View Post
    Try changing $db_selected=mysql_selected_db (DB_NAME, $link); to $db_selected=mysql_select_db (DB_NAME, $link);

    thanks i changed and now its showing no error, you are a great person, i wasted at least a week on these errors and you solved it in two messeges,

    thanks for your help


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