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xprt007

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  1. Does anyone here use PHP Eclipse & if so ... do you probably use XAMPP on your pc? I have tried to configure it, and indeed some few other open source IDEs, but end up failing, because of confusing configuration instructions or may be to be fair, that I find quite confusing. Some sites give the impression it's a few clicks away, after installing PHP eclipse + a few further steps, other give an endless & confusing list of steps & changes u are supposed to make to have it fully configured. Maguma has a free version which hardly needed any work on it to get it going, BUT they limit u to doing sites that have at most 20 files. Any help to configure PHP Eclipse or any other named Open Source PHP IDE for use at home with XAMPP (latest version) by someone who is relatively new to PHP? I have Xampp up & functioning at D:/Program Files/Xampp Where do I best install Eclipse (inside /xampp or htdocs? for example & so on up to when I can write & test php scripts) & what exactly do I do after downloading the PHP Eclipse plug in?
  2. Can anyone help? Please refer to that link & the posting below to get my dilemma. As said in that quote, I got the correct entry into the database of content from the four checked checkboxes. I discovered later, after my last posting that, strangely, that when I try to get that data from the database as a logged in user (using <? echo $_SESSION['Eigenschaften']; ?>), that particular field is blank, whereas all the other unrelated user information is shown. Is the "array" background of that data the problem? Anyway, how do I get it also shown in my members profile?? [!--quoteo(post=350704:date=Mar 1 2006, 06:57 PM:name=xprt007)--][div class=\'quotetop\']QUOTE(xprt007 @ Mar 1 2006, 06:57 PM) [snapback]350704[/snapback][/div][div class=\'quotemain\'][!--quotec--] I just noticed that I had made an un-intended double-posting. Apparently in the first posting, at [a href=\"http://www.phpfreaks.com/forums/index.php?showtopic=87173\" target=\"_blank\"]http://www.phpfreaks.com/forums/index.php?showtopic=87173[/a] I got a response that did work. Thanks again [/quote]
  3. I just noticed that I had made an un-intended double-posting. Apparently in the first posting, at [a href=\"http://www.phpfreaks.com/forums/index.php?showtopic=87173\" target=\"_blank\"]http://www.phpfreaks.com/forums/index.php?showtopic=87173[/a] I got a response that did work. Thanks again
  4. Thanks a great deal. In spite of making endless searches (googling on many PHP related sites) I'd not (yet) come across something like that. May be I have not been looking in the right places.
  5. Hi XenoPhage & kenrbnsn, Thank you for the very rapid responses. :) I first tried XenoPhage's suggestion... When I I tried to enter $varA together with other variables into database using [code]$sql = mysql_query("INSERT INTO customers(customerId,fname,sname,Eigenschaften,....) values (...,....,'$varA', .....[/code] I got "Arrays" again as database entry, whereas I got the 4 checked entries if I tried to echo it without entering it into database. I then tried kenrbnsn's suggestion, but may be got mixed up while trying to "customize" the code to fit what I've been using, ie making it use "INSERT" on a separate line. Since the customer entry is new & is just being created at that point ... the next line of code actually creates the ID [code]if($sql)                         {                                 $customerId = mysql_insert_id();[/code]... and so what I came up with, left the "Eigenschaften" column empty. Is it possible to integrate that entry under $sql, or at least to use INSERT and bearing in mind that ID is just being created then? I am relatively new to PHP as my questions may suggest, so please dont be too brief. Thanks again to both of you.
  6. Hi, I need some help urgently. [code]<td nowrap colspan="3"> Smoker:               <input type="checkbox" name="Eigenschaften[]" value="smoker" class="check">    Formular1 Fan:               <input type="checkbox" name="Eigenschaften[]" value="f1fan" class="check">    Soccer Fan:               <input type="checkbox" name="Eigenschaften[]" value="soccerfan" class="check">    Womanizer:               <input type="checkbox" name="Eigenschaften[]" value="womanizer" class="check">[/code] I want to get that input into a database. The problem is when I do it the normal way, ie $Eigenschaften = $_POST['Eigenschaften'] before entering it into database, I get the entry "array" in database. I must say I added the [] beside the variable name, after I was told arrays need to be used if more than one entry is automatically selected, even if one ticked more than one checkbox. That means I was always getting one entry into database. After endless googling, I have established one has got to use some form of arrays. There so many different suggestions, but none has led me to the desired result. An example: [code] if ($_POST['Eigenschaften'])         {         foreach($_POST['Eigenschaften'] as $element)                 {            echo "$element";         }     }[/code] That echos the entries I want. [!--coloro:#FF0000--][span style=\"color:#FF0000\"][!--/coloro--][Result after checking all the 4 boxes ==> smokerf1fansoccerfanwomanizer] [!--colorc--][/span][!--/colorc--] The problem is as soon as u replace say echo = "$element"; with say $varA = "$element"; to try & create a variable out of it, .. if one tries to enter it into database or echo it, it only returns one entry from the checkbox. The many examples I have seen after googling dozens of sites, only "echo" the result. I dont want that. I want all the checked entries together possibly stored in one variable like $varA ... that I can use for different things, before entering them in database. Does any one have an idea how to get the problem solved?
  7. Hi, I need some help urgently. [code]<td nowrap colspan="3"> Smoker:               <input type="checkbox" name="Eigenschaften[]" value="smoker" class="check">    Formular1 Fan:               <input type="checkbox" name="Eigenschaften[]" value="f1fan" class="check">    Soccer Fan:               <input type="checkbox" name="Eigenschaften[]" value="soccerfan" class="check">    Womanizer:               <input type="checkbox" name="Eigenschaften[]" value="womanizer" class="check">[/code] I want to get that input into a database. The problem is when I do it the normal way, ie $Eigenschaften = $_POST['Eigenschaften'] before entering it into database, I get the entry "array" in database. I must say I added the [] beside the variable name, after I was told arrays need to be used if more than one entry is automatically selected, even if one ticked more than one checkbox. That means I was always getting one entry into database. After endless googling, I have established one has got to use some form of arrays. There so many different suggestions, but none has led me to the desired result. An example: [code] if ($_POST['Eigenschaften'])         {         foreach($_POST['Eigenschaften'] as $element)                 {            echo "$element";         }     }[/code] That echos the entries I want. [!--coloro:#FF0000--][span style=\"color:#FF0000\"][!--/coloro--][Result after checking all the 4 boxes ==> smokerf1fansoccerfanwomanizer] [!--colorc--][/span][!--/colorc--] The problem is as soon as u replace say echo = "$element"; with say $varA = "$element"; to try & create a variable out of it, .. if one tries to enter it into database or echo it, it only returns one entry from the checkbox. The many examples I have seen after googling dozens of sites, only "echo" the result. I dont want that. I want all the checked entries together possibly stored in one variable like $varA ... that I can use for different things, before entering them in database. Does any one have an idea how to get the problem solved?
  8. Hello folks, I am new to php/mysql, as may be reflected by my question. I have a site where new members can register online & update their data and so on. That is functioning well. They are based on the tutorials http://www.phpfreaks.com/tutorials/78/0.php and http://& http://www.phpfreaks.com/tutorials/40/0.php I also have old members who registered in person, but need to update their data. I had to devise a log in mechanism some of the data in database, which after the initial log in, they update their other data & also get a password & account activation link emailed to them. After that I want them to log in as those whose registration was fully done online. Thats working too. The only problem, is some one in the database can use the specially designed form to repeatedly register. I have not managed to write a script that tells them their account has already been activated. I know I probably have to use some select query that checks that particular persons data & if they have \"Username\" and \"Password\" entered in the database, which they did not originally have, and the activation state at \"1\" and not at the default \"0\", but I dont know how to. How would the WHOLE query be? Something like: [php:1:510ac96c3d]<?php <?php session_start(); header(\"Cache-control: private\"); if (!$_SESSION[\'MUK_Names\']) { echo \"You\'re not logged in!\"; include(\"login2.html\"); exit(); } $MUK_Names = $_SESSION[\'MUK_Names\']; $userid = $_SESSION[\'userid\']; include (\"db.php\"); .... $sql_member_check = mysql_query(\"SELECT Username FROM Old_Students_MUK1 WHERE MUK_Names=\'$MUK_Names\' AND Registration_No=\'$Registration_No\' AND Year_of_Completion=\'$Year_of_Completion\'\"); $sql_encryptpass_check = mysql_query(\"SELECT encryptpass FROM Old_Students_MUK1 WHERE MUK_Names=\'$MUK_Names\' AND Registration_No=\'$Registration_No\' AND Year_of_Completion=\'$Year_of_Completion\'\"); $sql_activation_check = mysql_query(\"SELECT activated FROM Old_Students_MUK1 WHERE MUK_Names=\'$MUK_Names\' AND Registration_No=\'$Registration_No\' AND Year_of_Completion=\'$Year_of_Completion\'\"); ...... ?>[/php:1:510ac96c3d] and then what?? Or if I am wrong, how should it be? Under \"WHERE\", I included all those values, because the data is faulty with some almost similar entries in many fields which where causing trouble, until I did so. Can some one help? I also have at times or on some PCs an error message \"Stack Overflow at line 189\", when someone logs out. the logout.php page does not have that number of lines of code! What does it mean? The logout code looks like that: [php:1:510ac96c3d]<?php <? session_start(); if(!isset($_REQUEST[\'logmeout\'])){ echo \"<center>Are you sure you want to logout?</center><br />\"; echo \"<center><a href=logout.php?logmeout=true>Yes</a> | <a href=javascript:history.back()>No</a>\"; } else { session_destroy(); if(!session_is_registered(\'first_name\')){ echo \"<center><font color=red><strong>You are now logged out!</strong></font></center><br />\"; echo \"<center><strong>Login:</strong></center><br />\"; include \'login_form.html\'; } } ?> ?>[/php:1:510ac96c3d] Thanks, xprt007 :roll:
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