MySQL Query - some help needed

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Read Latest Entries..: (Post #11) by avalon on Feb 3 2005, 06:14 PM. (Line Breaks Removed)
CODE$sql="INSERT INTO `tablename` (`colname1`, `colname2`) VALUES ('value1', 'value2')";If you follow the above format, the insert query should work fine.You may have to define the columns name and data parallel to each other.
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MySQL Query - some help needed

panquetofobia
i dunno why this query doesn't function..

CODE

$sql="SELECT count(*) FROM participantes,sancionados WHERE participantes.No_Ficha = '$no_ficha' or sancionados.No_Ficha='$no_ficha'";
$query=($sql);



any suggestion???

Reply

avalon
CODE
$sql="SELECT count(*) FROM participantes, sancionados WHERE `participantes`.`No_Ficha` = '$no_ficha' AND `sancionados`.`No_Ficha`='$no_ficha'";
$query=$sql;


try the above code.

Reply

panquetofobia
doesn't function....

but thank u so much avalon wink.gif

Reply

avalon
Try this:-

CODE
$sql="SELECT count(*) FROM `participantes`, `sancionados` WHERE `participantes`.`No_Ficha` = '".$no_ficha."' AND `sancionados`.`No_Ficha`='".$no_ficha."'";
$query=$sql;


or this:-

CODE
$sql="SELECT count(*) FROM `participantes` WHERE `No_Ficha` = '".$no_ficha."'";
$query=$sql;


if it still doesn't not work, i will have to get back to you later today or tomorrow.

Reply

mastercomputers
Maybe you're not doing the query, you're just setting $query to $sql, you have to use $query = mysql_query($sql) or die("Maybe you were right?");

Other than this, I can't see a problem.


MC

Reply

miCRoSCoPiC^eaRthLinG
QUOTE
$sql="SELECT count(*) FROM participantes,sancionados WHERE participantes.No_Ficha = '$no_ficha' or sancionados.No_Ficha='$no_ficha'";
$query=($sql);


See from your query, I figure you are trying to match up the contents of the variable $no_ficha with the table columns participantes.No_Ficha and sancionados.No_Ficha. the query is perfectly fine and wouldn't even need the open-close quotes (``) around the table/column names as shown here.
QUOTE
`participantes`, `sancionados` WHERE `participantes`.`No_Ficha` = '".$no_ficha."' AND `sancionados`.`No_Ficha`='".$no_ficha."'";


My guess is where you are going wrong is the open-close quotes ('') around the $no_ficha.... if you want to match the contents of $no_ficha with your table columns, then you have to enclose $no_ficha in double quotes ("$no_ficha") for PHP to be able to parse it as a variable and extract and use its value. Between single quotes, $no_ficha is treated as a string and you query is trying to find a string called $no_ficha in your tables... Try with double quotes and see....

I think breaking up the query like this would help:

$sql="SELECT count(*) FROM participantes,sancionados WHERE participantes.No_Ficha =" . "$no_ficha" . "or sancionados.No_Ficha = " . "$no_ficha";

...
...
And next check your syntax and add what mastercomputers wrote...
QUOTE
Maybe you're not doing the query, you're just setting $query to $sql, you have to use $query = mysql_query($sql) or die("Maybe you were right?");
- coz it seems from your syntax that you aren't doing to query at all...

 

 

 


Reply

mastercomputers
I see where you're coming from microscopic^earthling and I did too pick this up.

The only thing if I can explain it to you is that the single quotes are inside double quotes, meaning they play no importance other than being single quotes in a double quoted string.

Because the double quoted string is used and there's no closing prior the single quotes, the variables will be processed by PHP as variables inside that double quoted string and not as if they were singluarly quoted.

e.g $double_quoted = " '$hello' "; produces 'the_element_contained_in_hello' with the quotes. If it were only $single_quoted = ' "$hello" '; produces only "$hello", with the double quotes around it and not what's contained in the variable $hello.

You need to use those quotes in those statements though, so never leave them off.

MC

Reply

miCRoSCoPiC^eaRthLinG
Righton smile.gif thanks for clearing it up...it got a little fuzzy seeing so many quotes flying around wink.gif

Reply

OpaQue
// Added Aliases just to avoid those big lengthy names
CODE

$sql="
SELECT count(*)
FROM participantes AS a, sancionados AS b
WHERE a.No_Ficha = '$no_ficha'
OR b.No_Ficha='$no_ficha';
";

$sql_result = mysql_query($sql,$connection) || die(mysql_error());

Check the Following,
The Column names are Case-sensitive and they are properly written.
If the things still dont work, Would you please put down the error statement reported by the "mysql_error()" function.

-OpaQue

Reply

panquetofobia
a part of my code...

CODE

$sqlcheck="SELECT count(*) FROM participantes AS a, sancionados AS b WHERE a.No_Ficha = '$no_ficha' OR b.No_Ficha='$no_ficha';";
  $querycheck=mysql_query($sqlcheck) || die(mysql_error());;

if(@mysql_result($querycheck,0,0)>0)
   {
    $merror.="<p style=\"color:#990000; font-weight:bold\">JUGADOR YA REGISTRADO &Oacute; SANCIONADO <img src=\"./imgs/error.gif\" alt=\"\"></p>";
   }
   else
   {
$sql="UPDATE equipos SET integrantes_restantes='$integrantesr' where Centro_de_Trabajo='$centro_trabajo' and Deporte='$deporte'";
$query=mysql_query($sql);
   $sql="INSERT INTO participantes VALUES (0,'$nombre','$ape1','$ape2','$no_ficha','$situacionc','$talla','$anios','$deporte','$rama','$centro_trabajo','$enc_reg','$fecha'";
   $query=mysql_query($sql);
   @$merror.="<p>Jugador ".$nombre." ".$ape1." ".$ape2." <strong>Registrado</strong>  <img src=\"./imgs/oki.gif\" alt=\"\"></p>";
   }  



the insert query doesn't function either sad.gif ...

thank u guys ...


greetings from México City

Reply

Latest Entries

avalon
CODE
$sql="INSERT INTO `tablename` (`colname1`, `colname2`) VALUES ('value1', 'value2')";


If you follow the above format, the insert query should work fine.
You may have to define the columns name and data parallel to each other.

Reply


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