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anybody help me php code

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reggiex
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Posted: Wed Feb 22nd, 2006 04:28 am

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anybody please help me with this:

how will i make this condition

$dlname is the variable for my txtbox set as POST

$dlname=$_POST['delete_lnametxt'];

$dfname=$_POST['delete_fnametxt'];

this is my condition

lname and fname are fields in my table in my database.i've already connect to the database. i want the value entered in the textbox(delete_lnametxt) to be check first in the database,if it exist then thats the time to execute my command..but i get an error with this condition..i think there's something wrong with the condition i put in my if statement..pls help..tnx.

if ((isset($dlname)==lname))&&(isset($dfname)==fname))){

ive already have a query and i think theres nothing wrong with it..i think whats wrong here is the condition i put in my if statement.

Last edited on Wed Feb 22nd, 2006 04:30 am by reggiex

klimaks
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Posted: Wed Feb 22nd, 2006 05:28 am

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if ((isset($dlname)==$lname)&&(isset($dfname)==$fname)){

reggiex
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Posted: Wed Feb 22nd, 2006 07:25 am

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ok tnx..ill try it..thnx a lot..

reggiex
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Posted: Thu Feb 23rd, 2006 05:06 am

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i tried it but still it wont work..i just want my value enter in the txtbox to be check if it exist in the database, and if it exist then execute so commands..can you give me an example code for that? just a simple example; im a beginner..ive already done adding a record in the database with php..and i want it to check the inputted text in the txtbox if it is in the database,if it exist then it wont add the same record..if you could give me a simple example much better..thanks

reggiex
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Posted: Thu Feb 23rd, 2006 05:10 am

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ive tried the example you gave me but i didnt work..ive tried adding a record to the database which is inputed from a textbox..i just want it to have a checking before it adds the record to the database..if the name you inputed in the textbox exist in the database then i wont add again the same record and it willl exit..im totally a beginner..if you have a simple example that is much better..tnx alot.

csg
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Posted: Thu Feb 23rd, 2006 05:16 am

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I found your problem. You are comparing your variables wrong. When you use the isset function it returns true or false (ie.. true for the variable exists or false it doesn't). So if the variable is true or not, when you compare the strings you are comparing a boolen value(true or false) and a string(text), and those are always false. What you'll want to do is check if the variable exists then compare the strings. Something like this:

if (isset($myvar)) {
    if ($myvar == $myvar2) {
       //item in database exists and matchs
    } else {
       //Item exists and does not match
} else {
    //Item does not exists in
}

reggiex
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Posted: Thu Feb 23rd, 2006 05:44 am

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in your example, what doest $myvar and $myvar2 represents? coz i did something like that already but it didnt work..

like:

$myvar=$_Post['txtboxname'];

$myvar2 should be the fieldname in your database;

(like example:my field name in the table inside my database is "lname")how will i compare that to $myvar(the inputed txt in the texbox)

and then i compared the two.

like if ($myvar==$myvar2){ -- if the inputed text is the same as a record in the database then......

csg
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Posted: Thu Feb 23rd, 2006 05:48 am

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Sorry about that. This was your example:

if ((isset($dlname)==$lname)&&(isset($dfname)==$fname)){

here is my

if (isset($dlname) && isset($dsname) {
    if ($dlname == $lname && $dfname == $fname) {
       //item in database exists and matchs
    } else {
       //Item exists and does not match
    }
} else {
    //Item does not exists in
}

reggiex
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Posted: Thu Feb 23rd, 2006 06:09 am

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ok thnx..ill try that..thanx a lot..

reggiex
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Posted: Thu Feb 23rd, 2006 06:11 am

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ok ill try that..thanx a lot...

reggiex
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Posted: Thu Feb 23rd, 2006 07:57 am

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its still dont work..by the way heres my code;

<?php
mysql_connect('localhost','root','april15') or die(mysql_error());
mysql_select_db('names')or die(mysql_error());

$l_name=$_POST['lnametxt'];
$f_name=$_POST['fnametxt'];

$trimmed=trim($l_name);//trim white spaces from stored variable so that the textbox will
$trimmed2=trim($f_name);//be filled up completely

if (isset($_POST['mybutton'])){
   //check if lnametxt and fnametxt are not blank
   if (($trimmed=="")||($trimmed2=="")){
    echo"please complete the form\n";
    exit;
    }
   if (isset($l_name)&& isset($f_name)){
    if ($l_name==$lname && $f_name==$fname){
     echo "You are currently registed in my database";
     exit;
    }
   }


//execute the query(to add)
else{
   mysql_query("INSERT INTO members(lname,fname) values('$l_name','$f_name')") or die(mysql_error());

   echo "<table width=750 border=2>Data Added\n";
   echo "<tr>";
   echo "<td>";
   echo "<h1>WELCOME $fname $lname</h1><br>";

   echo "</td>";
   echo "</tr>";

   echo "</table>";
   }




}

?>

reggiex
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Posted: Thu Feb 23rd, 2006 07:59 am

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it still dont work..by the way heres my code

<?php
mysql_connect('localhost','root','april15') or die(mysql_error());
mysql_select_db('names')or die(mysql_error());

$l_name=$_POST['lnametxt'];
$f_name=$_POST['fnametxt'];

$trimmed=trim($l_name);//trim white spaces from stored variable so that the textbox will
$trimmed2=trim($f_name);//be filled up completely

if (isset($_POST['mybutton'])){
   //check if lnametxt and fnametxt are not blank
   if (($trimmed=="")||($trimmed2=="")){
    echo"please complete the form\n";
    exit;
    }
   if (isset($l_name)&& isset($f_name)){
    if ($l_name==$lname && $f_name==$fname){
     echo "You are currently registed in my database";
     exit;
    }
   }


//execute the query(to add)
else{
   mysql_query("INSERT INTO members(lname,fname) values('$l_name','$f_name')") or die(mysql_error());

   echo "<table width=750 border=2>Data Added\n";
   echo "<tr>";
   echo "<td>";
   echo "<h1>WELCOME $fname $lname</h1><br>";

   echo "</td>";
   echo "</tr>";

   echo "</table>";
   }




}

?>

klimaks
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Posted: Thu Feb 23rd, 2006 08:13 am

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klimaks wrote:
if ((isset($dlname)==$lname)&&(isset($dfname)==$fname)){

My bad, should have been

if ((isset($dlname))&&($dlname==$lname)&&(isset($dfname))&&($dfname==$fname)) {

csg
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Posted: Thu Feb 23rd, 2006 08:14 am

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The variblaes $lname and $fname are not declared with anything. There for when you if statement is called php creates those variables, but with no value. So it will always be false. The other variables you are comparing are most likely from a user input, but then where are you wanting the other values from ($fname, $lname)?

reggiex
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Posted: Thu Feb 23rd, 2006 08:23 am

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the variables $fname and $lname are the fieldnames in my database..how will i do that?

csg
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Posted: Thu Feb 23rd, 2006 08:50 am

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Do you know how to get information from the database. Because you code doesn't have it. Example:

$result = mysql_query("SELECT lname, fname FROM members;") or die(mysql_error());

while ($row = mysql_fetch_array($result)) {
if ($l_name = $row['lname'] && ....) {
...
}

Once you have selected the information from the database it is in an array. You loop through the array to see if the user in any of the users. You have to change your code so that doesn't have mutilple results from the loop, for example. If you just add this code to your php file. It might add the user several times to the database.

Last edited on Thu Feb 23rd, 2006 08:56 am by csg

reggiex
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Posted: Fri Feb 24th, 2006 12:43 am

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thanx..i havent tried asking information from the database yet..im just a beginner..ive only made adding info into the database and deleting..ill try searching and displaying later..thanx a lot...

mark-o-matic
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Posted: Mon Apr 17th, 2006 10:44 am

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csg wrote: Do you know how to get information from the database. Because you code doesn't have it. Example:

$result = mysql_query("SELECT lname, fname FROM members;") or die(mysql_error());

while ($row = mysql_fetch_array($result)) {
if ($l_name = $row['lname'] && ....) {
...
}

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