iRush
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Everything posted by iRush
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Hey guys i have a few question with this login script im trying to make. Ok so first im using php myadmin and i've created a simple login here that works fine but i want to tweak it so when i login i can login to a specific site where i've created tables and stuff. Is there anyone out there that can help me im a little stumped on how to do this. Thanks in regards. Here are my two scripts im using show_login.html <html> <head> <title>Login Form</title> <h1>Login to Authorized Area</h1> <form method= "post" action= "login.php"> <br> Username: <input type= "text" name= "username"> <br> <br> Password: <input type= "password" name= "password"> <br> <p> <input type= "submit" name= "submit" value= "Login" </p><br> </html> </form> login.php <? if ((!$_POST['username']) || (!$_POST['password'])) { header("Location: show_login.html"); exit; } $db_name="ryan"; $table_name="auth_users4"; $connection=@mysql_connect("localhost","RyanH","test1234")or die(mysql_error()); $db =@mysql_select_db($db_name, $connection)or die(mysql_error()); $sql= "SELECT * FROM $table_name WHERE username ='$_POST[username]' AND password = password('$_POST[password]')"; $result = @mysql_query($sql,$connection) or die(mysql_error()); $num=mysql_num_rows($result); if ($num != 0) { echo "<P>Congratulations, you're authorized!</p>"; }else{ echo "<P>You are not authorized to use this site</p>"; echo'<br/><a href="show_login.html">Try Again</a>'; exit; } ?>
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anyone have any ideas on how i would incorporate the table into my code
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This is my code <?PHP // please add login and pass here// $host = "localhost"; $login = "root" ; $pass = ""; mysql_connect("$host","$login","$pass") OR DIE ("There is a problem with the system. Please notify your system administrator." .mysql_error()); //Seems in this case we can use a general call $connection = mysql_connect("$host","$login","$pass") or die(mysql_error()); $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } //Start Create DB// IF (isset($_POST['result'])){ $database=$_POST['database']; $sql="CREATE DATABASE $database "; $result = mysql_query($sql,$connection) or die(mysql_error()); echo "Database $database has been added"; } IF (isset($_POST['delete'])){ $db=$_POST['db']; $query=mysql_query("DROP DATABASE $db"); echo "Database $db has been deleted"; } ?> <html> <head> <title>MySQL Databases</title> </head> <body> <p><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> <?PHP //print_r($_POST); ?> <form action="pretask.php" method="post"> <select name="db"> <?PHP $db_list = mysql_list_dbs($connection); while ($row = mysql_fetch_object($db_list)) { //Here you are listing anything that should not be included if ($row->Database!="information_schema" && $row->Database!="mysql" && $row->Database!="phpmyadmin"){ echo "<option value=\"".$row->Database."\">".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="Delete"/> </form> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body>
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So if i post the code i already have that has a list of databases and which also contains me being able to add and remove databases from a list can you help me out as to where i should implant the code so i can do it with a table instead?
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I need some help on just making a plain and simple php table with rows and columns..then i need to beable to add in data from a text box form and submit button. So say i type in the text box test1 then it should put it in the table at row 1 column 1. Any ideas or help would be great to get me started
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$member=$_POST['Name']; That should be member inside instead of name im guessing
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I just want to put this list into a table a simple list from my localhost into a table
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Anyone have any ideas
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Now that i have my code that shows my list on the localhost I want to try and put this into a table and beable to add a delete from the table. Here's my code any help would be much appreciated. <?PHP // please add login and pass here// $host = "localhost"; $login = "root" ; $pass = ""; mysql_connect("$host","$login","$pass") OR DIE ("There is a problem with the system. Please notify your system administrator." .mysql_error()); //Seems in this case we can use a general call $connection = mysql_connect("$host","$login","$pass") or die(mysql_error()); $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } //Start Create DB// IF (isset($_POST['result'])){ $database=$_POST['database']; $sql="CREATE DATABASE $database "; $result = mysql_query($sql,$connection) or die(mysql_error()); echo "Database $database has been added"; } IF (isset($_POST['delete'])){ $db=$_POST['db']; $query=mysql_query("DROP DATABASE $db"); echo "Database $db has been deleted"; } ?> <html> <head> <title>MySQL Databases</title> </head> <body> <p><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> <?PHP //print_r($_POST); ?> <form action="pretask.php" method="post"> <select name="db"> <?PHP $db_list = mysql_list_dbs($connection); while ($row = mysql_fetch_object($db_list)) { //Here you are listing anything that should not be included if ($row->Database!="information_schema" && $row->Database!="mysql" && $row->Database!="phpmyadmin"){ echo "<option value=\"".$row->Database."\">".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="Delete"/> </form> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body>
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I'll try it tomorrow..will the database follow over to the next page as well? Thanks bud for all your help so far..you've been great!
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Didnt change anything..still have to refresh. This is a tricky one to figure out
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Ok the code you gave me works mint..but when i create and delete databases i have to refresh the page once or more to see the results. Any ideas?
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Do I need this in the code while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; $db_list .="</ul>" or this <? echo "$db_list"; ?>
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I need my original list to show up somehow but no databases are showing up. Also the delete dropdown list isnt showing databases either probably because none are being show
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^^^^^^^^^ anyone?
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Did you scroll down through the whole code...i added what you told my to..to my code that i already had started
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Yeah here is my full code and i have a create script right above that im not sure is 100% right or not.. <? $connection = @mysql_connect("localhost", "root", "") or die(mysql_error());; $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } $db_list .="</ul>"; ?> <HTML> <HEAD> <TITLE>MySQL Databases</TITLE> </HEAD> <P><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> </BODY> </HTML> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body> </html> <?php $database=$_POST['database']; $connection = @mysql_connect("localhost","root","") or die(mysql_error()); $sql="CREATE DATABASE $database "; $result = @mysql_query($sql,$connection) or die(mysql_error()); if (isset($_POST['result'])) { Results: echo "Database $database has been added"; } mysql_close($connection) ?> <?PHP print_r($_POST); ?> <form action="pretask.php" method="post"> <select name="db"> </html> <?PHP $connection = mysql_connect("localhost","root","") or die(mysql_error());; $db_list = mysql_list_dbs($connection); while ($row = mysql_fetch_object($db_list)) { if ($row->Database!="information_schema" && $row->Database!="mysql" && $row->Database!="phpmyadmin"){ echo "<option value=\"".$row->Database."\">".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="delete"/> </form>
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It says this now Database testdb6 has been added Array ( [database] => testdb6 [result] => Create ) And it still doesn't delete anything from the list. Thank you so much for helping out this far though I appreciate it alot! I just want to figure this out
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Do i have to change the "echo option value at the bottom?
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Where would i add print_r($POST) if you dont mind me asking
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Yeah i did and when i select one and hit delete nothing happens it doesnt delete it from the list of databases. It comes up with that error though in the above post.
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Ok this helped me out alot this is my updated code with what you told me to add.. but this error came up..i tried choosing one of the databases i hit the delete button but it did not delete it. Here is the error: Notice: Undefined index: database in C:\xampp\htdocs\pretask.php on line 42 <? $connection = @mysql_connect("localhost", "root", "") or die(mysql_error());; $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } $db_list .="</ul>"; ?> <HTML> <HEAD> <TITLE>MySQL Databases</TITLE> </HEAD> <P><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> </BODY> </HTML> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body> </html> <?php $database=$_POST['database']; $connection = @mysql_connect("localhost","root","") or die(mysql_error()); $sql="CREATE DATABASE $database "; $result = @mysql_query($sql,$connection) or die(mysql_error()); if (isset($_POST['result'])) { Results: echo "Database $database has been added"; } mysql_close($connection) ?> <html> <body> <form action="pretask.php" method="post"> <select> <?PHP $connection = mysql_connect("localhost","root","") or die(mysql_error());; $db_list = mysql_list_dbs($connection); while ($row = mysql_fetch_object($db_list)) { if ($row->Database!="information_schema" && $row->Database!="mysql" && $row->Database!="phpmyadmin"){ echo "<option>".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="delete"/> </form>
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Is there a shortcut instead of writing out all the databases in my if statement? Would it be easier to make another text field and use a delete "submit button" to delete the databases?
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I have this for my drop down list but how do i actually show all the databases i have to use the variable $db_list somewhere not too sure though. It just shows $db_list and not the actual list is there another way to make a drop down list instead of this way? <html> <body> <form action="pretask.php" method="post"> <select> <option>$db_list</option> </select> <input type="submit" name="delete" value="delete"/> </form> </body> </html>
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Where would i put this in my code and i would have to make another submit button and add a value to it just like i did for the create button.