619 Posted Topics
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| | For displaying it back you should use mysql db. So once image is uploaded on server i.e. @ echo "<h1>File Uploaded Successfully! Try again!</h1>"; $newname will be saved in table. After that using select query that $newname will be fetched and shown in image src tag. |
 | You can go for "cronjob" for this task.hope you know about cronjob. logic: create one table in database.Where fields will be : id, filePath, deleteDate, status set cronjob task daily with one php file. this php file will read table and delete file from filePath on deleteDate and update status. |
| | use any good jquery gallery. check this : [URL="http://www.1stwebdesigner.com/design/fresh-jquery-image-gallery-display-solutions/"]Jquery Lists[/URL] Once you download script and demo is working in you pc, replace static images to dynamic images. Dynamic images will comes from select query results.  |
| | Its simple as you do code for insert same thing should be there for update. Like insert query here you will have update query. [CODE] <html> <head> <title>Eastern Mediterranean University-Student`s ID</title> </head> <body> <h3>Edit Student</h3> <?php $num=$_POST['num']; @ $db = new mysqli('localhost', 'root', 'ilyas1987', 'dau'); if (mysqli_connect_errno()) { echo 'Error: … |
| | also post what you are getting as a result in print_r($productList2) and print_r($totalProductList) |
| | Re: creating excel echo $query_string; exit; Check sql query on your domain and execute in sql window, what you are getting? |
| | put single qoute for LastName as it is varchar. [CODE] global $conn,$dal; $customerID = mysql_insert_id(); $strSQLInsert = "insert into Orders (CustomerID, LastName) values (".$customerID.", '".$values["LastName"]."')"; db_exec($strSQLInsert,$conn); $rs = $dal->Orders->Query("CustomerID=".$customerID." and LastName='".$values["LastName"],"'"); $data = db_fetch_array($rs); $idvalue = $data["ID"]; header("Location: Orders_edit.php?editid1=".$idvalue); exit(); [/CODE] |
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| | [CODE] $query5="INSERT INTO $clean_title (recnum,lname,fname,first,credit) VALUES ('$record_number1','$last_name','$first_name','$new','$credit')"; mysql_query($query5) or die(mysql_error()); [/CODE] there is no php code to execute query in your given code. same for $query4. |
 | Try this : [CODE] <?php $new = htmlspecialchars("'The <font> tag is awesome'", ENT_QUOTES); echo $new; ?> [/CODE] |
| | you should create object of class system_class. then only all function of that class can be used. |
| | Re: php checkboxes check this thread. It will help you. [URL="http://www.daniweb.com/forums/thread98678.html"]http://www.daniweb.com/forums/thread98678.html[/URL] |
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 | first of all save all uploaded files to your website's folder. Then give url of that location in img src. $aspirant_photo = url of uploaded image $msg .= "Aspirant's Offer:<img src='".$aspirant_photo."' /> \n"; |
| | I think your y.innerHTML string is broken, try to add back slashed before single qoute e.g. beside window.status= [CODE] y.innerHTML='<input type = "text" name = "day"><img src="images/calendar.gif" name="imgCalendar" width="34" height="21" border="0" alt=""><a href="javascript<b></b>: void(0);" onmouseover="if (timeoutId) clearTimeout(timeoutId);window.status='Show Calendar';return true;" onmouseout="if (timeoutDelay) calendarTimeout();window.status='';" onclick="g_Calendar.show(event,'frm.day',true,'yyyy-mm-dd'); return false;"></a> '; [/CODE] |
 | [CODE] <form action="http://www.google.com/translate" > <script language="JavaScript"> <!-- document.write ("<input name=u value="+location.href+" type=hidden>") // --> </script> <input name="hl" value="en" type="hidden"> <input name="ie" value="UTF8" type="hidden"> <input name="langpair" value="" type="hidden"> <input name="langpair" value="en|fr" title="French" src= "http://photos1.blogger.com/img/43/1633/320/13539949_e76af75976.jpg" onclick="this.form.langpair.value=this.value" height="20" type="image" width="30"> <input name="langpair" value="en|de" title="German" src= "http://photos1.blogger.com/img/43/1633/320/13539933_041ca1eda2.jpg" onclick="this.form.langpair.value=this.value" height="20" type="image" width="30"> <input name="langpair" value="en|it" …  |
| | in libmail.php there is function Send(). First of all echo all member of class Mail folowing exit. If all members are Ok, then something is missing with your mail() function.trace mail function. check this below testing email code. [CODE] <?php $to = 'youremailaddress@example.com'; $subject = 'the subject'; $message = 'hello'; … |
| | noneed to redirect1.php page. FORM tag should be outside of for loop. [CODE] <?php if(isset($_REQUEST['Submit']) && isset($_REQUEST['name'])) { $allname = $_REQUEST['name']; for($i=0;$i<count($allname);$i++) { $q="insert into user(id,name) values('','".$allname[$i]."')"; mysql_query($q); } header("location:initial.html"); exit; } $val = $_POST['yourname']; ?> <form action="" method="post"> <?php for ( $i = 0; $i < $val ; $i++){ … |
| | Try this code. [CODE] <?php while($rows=mysql_fetch_array($result)) { $style = ''; if($rows['designation']=="Teamleader") $style = 'style="color:#0000FF;"'; if($rows['designation']=="Manager") $style = 'style="color:#cc0000;"'; echo "<option value=".$rows['username']." ".$style.">".$rows['username']."</option>"; } ?> [/CODE] |
| | your $sql query and return resultset is same for both foreach.so it is little bit ambiguous. Post ur requirement, so anyone can give you logic. |
 | I have made this code.. Check this out. [CODE] <table> <?php require "conetion.php"; $query = "SELECT * FROM forms WHERE active = '1' ORDER BY Name ASC"; $result=mysql_query ($query); $total_rows = mysql_num_rows($result); $count = 1; while($row = mysql_fetch_array($result)) { $all[] = $row['Name']; } $halfCount = ceil(count($all)/2) ; $ptr1 = 0; … |
 | Re: sql error What ankit means, [CODE] function fetch_row($resource) { $result = mysql_fetch_row($resource); return $result; } function fetch_array($resource) { $result = mysql_fetch_array($resource); return $result; } $q="select * from user"; $res = mysql_query($q); $output = fetch_row($res ); [/CODE] Firstly check your sql query.Is it right or not? and return of mysql_query should be passed … |
| | Here you have echo displayUsers(); meance whatever will be return from that function will be echoed. Your whole function is concating $display and returning it. But in if statement you have directly make echo. [CODE] //THIS IF STATEMENT DISPLAYS WHAT I NEED PERFECTLY if (is_null($avatar)) { echo "<img src=\"avatars/thumbs/default.jpg\"class=\"thumb\" alt=\"Avatar\" … |
| | Try this code.instead of nine conditions. Just for your review. Here you have 3 fields and u r using 3*3=9 if-else, if you will have four fields then?It is not good practice for coding [CODE] if(!empty($_POST['Alumnos'])) { $sql = " (SELECT email FROM alumnos ORDER BY nombre) "; } if(!empty($_POST['Maestros'])) … |
| | [CODE] <script language="javascript"> function getXMLHTTP() { //fuction to return the xml http object var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ req = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getSubCat(strURL) { var req = getXMLHTTP(); // fuction … |
 | you mean if there are three option tag then u want to store all in db? not the selected one? Give example here what you want... |
| | Re: Need help please For checking word is there or not: Here is code: [CODE] <? $file='path/to/file.txt'; $data=file_get_contents($file); $findme = 'some word'; $pos = strpos($data, $findme); // Note the use of ===. Simply == would not work as expected // because the position of 'a' was the 0th (first) character. if ($pos === false) … |
| | <chart caption='COST OF TURNOVER' subCaption='' showPercentValues='1' pieSliceDepth='30' showBorder='0'> is behaving as html tags... So it is coming in output but you wont be able to see it.It can be seen in view source. You want to disply it as a text?? Try this: [CODE] <?php $str = "<chart caption='COST OF … |
| | Hi all, I am having one music website. where admin upload songs and user can download it after paypal payment. Is there any way admin can provides rights to this song? like DRM rights in original cds. Help !!!!!!!!!!!!!!!!!! |
| | means you want to know how session works?? you can check session and if not found send user to login page. [CODE] if( !isset($_SESSION['admin_login']) && !$_SESSION['admin_login']) { header("location:login.php"); } [/CODE] during login you can set settion value like below. [CODE] session_start(); include("connexion.php"); if (isset($_POST['login'])and isset($_POST['password'])) { extract($_POST); if (empty($login) or … |
| | when user is login save userid in session from that point only. so you can directly use it, $query ="SELECT * FROM databasetable where userId = ".$_session['userId']; $result = ...fetch above query.... $oppositeSql = "SELECT * FROM databasetable where gender != ".$result[0]['gender']; this will give opposite gender's list |
| | I have uploaded all my files on the Apache/2.2.3 (Red Hat) live server. There is htaccess file on the root of website for converting php to html. But it is not working and getting "Internal Server Error". If i blank that htaccess code then it works. I don't know what …  |
| | This problem is on AWS linux instance,in local its working perfact. Here is my flow: this file is:test.php [CODE] require_once('system/config.php'); $_SESSION['name']='test'; echo 'session -name- is:'.$_SESSION['name']; exit; [/CODE] Expected Output: session -name- is:test (its coming.OK) here i have set session value to test. so in page session is now live and … |
| | check the db entry for 'Dyl' and for login echo sql query for line number 18. Execute that echo sql query in phpmyadmin.does it works? What happends... check and post back. |
| | Need selected in if else for each option tag. [CODE] <td> <label>Billing Month <span class="small">Like September-2010</span> </label></td> <td> <? $year = date("Y") ; if(isset($_REQUEST['submit']))// if submit btn is pressed { $b = $_REQUEST['b']; } ?> <select name="b" class="billmonth"/ > <option value=<? echo "January-".$year; ?> <?=($b == "January-".$year)?'selected="selected"':'';?> ><? echo "January-".$year; … |
| | You can also use php html parser. Which will fetch title tag, metadata keyword and description from URL. So in index page you can display this fetched title and metadata dynamically from html file, without saving or storing any information. |
| | post ur code where checkbox value is taken frm form. |
| | Re: php & sql Not sure but you can execute these two query separately. First insert query then execute it and then select query and again execute. Its a trial & error. |
| | What is need for inputbox, do you want to save it in db? If yes then here there is no any uniqueid's concept. |
| | this will guide you, [URL="http://devzone.zend.com/article/3336"]http://devzone.zend.com/article/3336[/URL] |
 | After importing csv, run this query in sql tab of phpmyadmin. [CODE] alter table tbltest add id int(11) primary key auto_increment not null first; [/CODE] Here tbltest is table name and id is auto incremented field. |
| | Kindly check whether $_SESSION['loggedIn'] session is flushed away during logout. And what is exact steps and which is not coming in your page bcz of session error? |
| | Re: Display Notes Hi andydeans, Just fix width of the container which is having this note table. (i.e. parent tag of <table width="50%">) not in percentage but in pixel. |
| | Re: PHP classes If index.php and add.php, edit.php... etc are different php pages then you have to manually create object instance for each page. You can also do all these things like add, edit, delete, search in only one page. which can use one single instance. |
| | Re: Form Prpblem Why there is two form tag???? Use only once after <body><form...> and complete it before</form></body> And --> <select name="category" onChange="submit()"> What is need of onChange submit??? |
| | Hey, ur actpopup funstion is unclear. Here i have define function. hope it works. [CODE] <script language="javascript"> function actpopup(act_id) { var url = "serv_act.php?act_id="+act_id; var opt = "height=175, width=300, location=no, scrollbars=no,menubars=no,toolbars=no,resizable=yes,left=50"; var name = "popup"; window.open(url,name,opt); } </script> [/CODE] |
| | Re: Drop Down Menu Hi, [CODE]<?php if(isset($_REQUEST['send'])) { $mydrodown = $_REQUEST['mydrodown']; echo "You have selected ".$mydrodown." from drop down."; exit; } ?> <form id="form1" name="form1" method="post" action=""> <select name="mydrodown"> <option value="computer memory">computer memory</option> <option value="computer parts">computer parts</option> <option value="computer peripherals">computer peripherals</option> </select> <input name="send" value="send" type="submit" /> </form>[/CODE] Here i have given example of … |
| | [CODE] <? $currTime = date('H:i'); if($currTime >= "08:00" && $currTime <= "20:00") { ?><a href="#">Time is between 8:00 am - 8:00 pm.</a><? } ?> [/CODE] Try this code.  |
| | Hi, try this code. [CODE] $bedrooms=$_POST['bedrooms']; $bathrooms=$_POST['bathrooms']; $whereStr = ' 1=1 '; if(trim($bedrooms)!='') $whereStr.= " AND bedrooms like '$bedrooms'" ; if(trim($bathrooms)!='') $whereStr.= " AND bathrooms like '$bathrooms'" ; $query= "SELECT * FROM inventory WHERE ".$whereStr." ORDER BY model_name ASC"; [/CODE] |
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