help me in this code correct it plz....

medo

hi friends

i wrote this code to add joke and rerieve it from database

<html dir=rtl>
<?php
if (isset($addjoke)):?>
<FORM ACTION-"<?php echo($PHP_SELF);?>"
 METHOD=POST><p>write your joke here to added it<BR>
<TEXTAREA NAME -"joketext" rows-10 cols-40 warp rows="9" cols="42">
</TEXTAREA><BR><INPUT TYPE = SUBMIT NAME ="subnitjoke" 
value-"ÇÖÝ"></form>
<?php else :
$dbcnx =@mysql_connect("localhost","root","");
If (!dbcnx) {
echo ("<p>can't connect to database</p>");
exit();
}
If(!mysql_select_db("jokes"))
{
echo ("<p>can't connect to database</p>");
exit();
}
if ("ÇÖÝ"==$subnitjoke){
$sql - "insert into jokes set joketext='$joketext',jukedate=curdate()";
if(mysql_query($sql)) {
echo ("<p>the joke added</p>");
}else {
echo(maysql_error());
}
}
echo("<p> this all the joke in the database</p>");
$result=mysql_query("select * from jokes");
if(!result) {
echo(maysql_error());
exit();
}
while($row = mysql_fetch_array($result)){
echo("<p>" .$row["joketext"] ."</p>");
}
echo("<p><A HREF-'$PHP_SELF?addjoke-1'>"." add joke</A></p>");
endif
?>

tomhath

It would help if you had said what errors you were getting. Try this, it should be close to what you want:

<html dir=rtl> 
<?php 
if (!isset($_POST['subnitjoke'])): ?> 
 <FORM ACTION-"<?php echo($_SERVER['PHP_SELF'] );?>" METHOD=POST>
   <p>write your joke here to added it<BR> 
   <TEXTAREA NAME="joketext" rows="9" cols="42"> </TEXTAREA>
   <BR><INPUT TYPE=SUBMIT NAME="subnitjoke" value="ÇÖÝ">
 </form> 
<?php  

else : 
  $dbcnx = @mysql_connect("localhost","root",""); 
  If (!$dbcnx) { 
    echo ("<p>can't connect to database</p>"); 
    exit(); 
  } 
  If(!mysql_select_db("jokes")) 
  { 
    echo ("<p>can't connect to database</p>"); 
    exit(); 
  } 
  if ("ÇÖÝ" == $_POST['subnitjoke']){ 
    $sql = "insert into jokes set joketext='" . $_POST['joketext'] . "',jukedate='curdate()'"; 
    if(mysql_query($sql)) { 
      echo ("<p>the joke added</p>"); 
    }else { 
      echo(mysql_error()); 
    } 
  } 
  echo("<p> this all the joke in the database</p>"); 
  $result=mysql_query("select * from jokes"); 
  if (!$result) { 
    echo(mysql_error()); 
    exit(); 
  } 
  while($row = mysql_fetch_array($result)){ 
    echo("<p>" . $row["joketext"] ."</p>"); 
  } 
  echo("<p><A HREF='" . $_SERVER['PHP_SELF'] . "'> add joke</A></p>");   

endif
?>

medo

thanks a alot 🙂

when i run the code in my server i faced this problem

(Unknown column 'jukedate' in 'field list')

Unknown column 'jukedate' in 'field list'

this all the joke in the database

joke1

joke2

add joke

tomhath

Did you mean "jokedate" instead of "jukedate"?

$sql = "insert into jokes set joketext='" . $_POST['joketext'] . "',jukedate='curdate()'";

medo

my table structure is :

CREATE TABLE jokes (
joketext varchar(100) NOT NULL default '',
,jukedate date default NULL

) TYPE=MyISAM;

when i run the your code the resylt is :

Unknown column 'jukedate' in 'field list'

this all the joke in the database

joke1

joke2

add joke

i dont now where is the problem ?

medo

hi agin

is this code right

$sql = "insert into jokes set joketext='" . $_POST['joketext'] . "',jukedate='curdate()'";

becuse when i need insert value in the table use i shoud use this code

insert into tablename values(v1,v2,......)

please correct the error

laserlight

If I remember correctly the INSERT...SET syntax should work too.
If you're not certain, just change to the syntax you're more familiar with.

I think you have to remove the single quotes in:
jukedate='curdate()'

medo

i removed the single quotes in the curdate() the problen still

the result for run the code is

Unknown column 'jukedate' in 'field list'
this all the joke in the database

joke1

joke2

add joke

this is the code

<html dir=rtl> 
<?php  

if (!isset($_POST['subnitjoke'])): ?>  

<FORM ACTION-"<?php echo($_SERVER['PHP_SELF'] );?>" METHOD=POST> 
   <p>write your joke here to added it<BR> 
   <TEXTAREA NAME="joketext" rows="9" cols="42"> </TEXTAREA> 
   <BR><INPUT TYPE=SUBMIT NAME="subnitjoke" value="add"> 
</form> 
<?php   

else :  

  $dbcnx = @mysql_connect("localhost","dbuname","pass");  

  If (!$dbcnx) {  

    echo ("<p>can't connect to database99</p>");  

    exit();
	  }  

  If(!mysql_select_db("mydb"))  

  {  

    echo ("<p>can't connect to database</p>");  

    exit();  

  }  

  if ("add" == $_POST['subnitjoke']){  

$sql = "insert into jokes set joketext='" . $_POST['joketext'] . "',jukedate=curdate()";

if(mysql_query($sql)) {  
  echo ("<p>the joke added</p>");  
}else {  
  echo(mysql_error());  
}  
  }  

  echo("<p> this all the joke in the database</p>");  

  $result=mysql_query("select * from jokes");  

  if (!$result) {  

    echo(mysql_error());  

    exit();  

  }  

  while($row = mysql_fetch_array($result)){  

    echo("<p>" . $row["joketext"] ."</p>");  

  }  

  echo("<p><A HREF='" . $_SERVER['PHP_SELF'] . "'> add joke</A></p>");    

endif 
?>

tomhath

The message coming from mysql_error is probably right. There is no column "jukedate" in your table. So take a close look at the table column names (maybe with either phpmyadmin or by doing a "select * from jokes").

From what you posted, I'd say you created the table incorrectly, look very closely at the code. There's an extra comma ',jukedate'

CREATE TABLE `jokes` ( 
`joketext` varchar(100) NOT NULL default '', 
`,jukedate` date default NULL 
) TYPE=MyISAM;

medo

the table created sucessfuly and i add data in it before and but when i run the code i faced the error 🙁

if you can write new code to add record to the table in the data base and retrieve the data from it u solve the problem>🙂

tomhath

Try dropping the table, then create it again with this. The code you already have should work:

CREATE TABLE `jokes` (  

`joketext` varchar(100) NOT NULL default ,  

`jukedate` date default NULL  

) TYPE=MyISAM;

medo

hi friend

the code doesnot work becuse their is no values for the defualt value in the juletext

CREATE TABLE jokes (

joketext varchar(100) NOT NULL default ,

jukedate date default NULL

) TYPE=MyISAM;

tomhath

Take off the "default". That will also keep you from getting blank rows in the database.

CREATE TABLE `jokes` (   

`joketext` varchar(100) NOT NULL,   

`jukedate` date default NULL   

) TYPE=MyISAM;