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PHP - Store/retrieve Html Code With Mysql
Hello guys,
I am a new programmer and i am building a new website. I would like to give me your advice for the following problem: I want to build a webpage, in which there will be a <div id="book-content"> ...............</div> part. Inside this div i would like to dynamically display pages from a book. Each page will have text, scripting code blocks, blocks with the output of each scripting code, and images. There will be a bar on the left of the webpage in which the user can select which page of the book he wants to load. For example...My web page will look something like this... Code: [Select] <HTML> <HEAD> </HEAD> <BODY> <DIV ID="PAGE-HEADER"> //LOGO OF THE WEB PAGE </DIV> <DIV ID="PAGE-MENU"> //PAGE MENU </DIV> <DIV ID="BOOK-INDEX" WITH FLOAT:LEFT > //HERE A WILL SHOW THE CHAPTERS OF THE BOOK AND THE PAGES OF EACH CHAPTER </DIV> <DIV ID="BOOK-PAGE"> //THE PAGE SELECTED FROM THE PREVIOUS 'BOOK-INDEX' MENU WILL BE DISPLAYED HERE WITH A MYSQL QUERY ****** </DIV> </BODY> </HTML> Then in a mysql database, i would like to have records with a text field, that will contain for example the followng: <h1> Chapter 1: bla bla </h1> <p> in this chapter we will speak about bla bla bla.... </p> <div id="code"> int main() { int x,y; x=2; y=3; x=x+y; } </div> <p> this will outpout the following:</p> <div id="code output"> x=5! </div> I would like to get this html code from the database, and then show it in the <div id=BOOK-PAGE> div. But i dont want to use php eval(). Also, if i store the code to a file and then include it, i will have too may files(equal to the book's number of pages etc 100). Any ideas? Similar TutorialsI m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php include("db.php"); $query=mysql_query("select * from table where id='3' "); $row=mysql_fetch_array($query); $r=$row['image']; header("content-type:image"); echo $r; ?>" i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance... Hi, Im trying to retrieve HTML from a mysql database but nothing i've tried seems to work. The HTML im trying to retrieve is an iframe with a link and styles (code from amazon associates). Im trying to display links to specific products on amazon from the product page on my site. All data about the product is retrieved from the database so i have code to select the amazon link row in my database table but i cant get it to display. It says the html isnt a string so i cant echo it, fair enough. I have tried using the following code: Code: [Select] $get_buylink_sql = "SELECT mobo_buylink FROM mobo WHERE mobo_id = $mobo_id"; $get_buylink_res = mysqli_query($mysqli, $get_buylink_sql) or die(mysqli_error($mysqli)); while ($buylink = mysqli_fetch_array($get_buylink_res)) { echo"<iframe src=\"".$buylink."\" style=\"width:120px;height:240px;\" scrolling=\"no\" marginwidth=\"0\" marginheight=\"0\" frameborder=\"0\"></iframe>"; } mysqli_free_result($get_buylink_res); mysqli_close($mysqli);I have also tried putting the whole iframe code in the database which didnt work either. The mobo_id variable works fine for retrieving the rest of the data and i need to get the amazon link from the same record. I hope i've put this in a way you can understand, but if not i'll try and explain better and give you a link to my site if needed. Thanks, Alex Hi, My output in html is : Code: [Select] <uL><li id="B1"></li> <li id="B2"></li> <li id="B3"></li> <li id="B4"></li> <li id="B5"></li> <li id="B6"></li> <li id="B7"></li> <li id="B8"></li> <li id="B9"></li> <li id="B10"></li> <li id="B11"></li> <li id="B12" class="active"></li> <li id="B13" class="no"></li> <li id="B14" class="no"></li> <li id="B15" class="no"></li> <li id="B16" class="no"></li> <li id="B17" class="no"></li> <li id="B18" class="no"></li> <li id="B19" class="no"></li> <li id="B20" class="no"></li> </ul> If MySQL query result is equal to `6`, then `<li>` tag with `id` equal to "`B12`" should have class "`active`". All the `<li>` elements occuring after this active element should have class "`no`". This shows images of horizontal rating between `0` and `10` and between 0.5 Example : 0 .5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 In the example above elements from `0` to `5` will be blue, element `6` will be white and elements from `7` to `10` are black. How could I generate this using PHP and/or MySQL? Thanks, XXXX chirstmas XXXX Hello dear friends, If i've database table (my_table) with one field (text) and i've created form with textarea Code: [Select] <form action="#"> <textarea name="text" id="text"></textarea> <input type="submit" name="Submit" value="Submit"> </form> I can add any text inside it and save it in my database table but when i add html codes it also store it perfect but only 2 codes is not the closing of form and textarea </form> and </textarea> it not stored and automatically not saved Why ! i mean if i paste inside the above form - textarea the following code for example Code: [Select] <form> <textarea>blah</textarea> </form> it will store it as Code: [Select] <form> <textarea>blah and will not store </form> and </textarea> how then i can add html codes with textarea and form inside it without conflict thanks Hello people, I have created an application in Code Charge where my site users can login and play games online.The game code is run via flash and html. The Game has 5 stages and from a page in my application called transfertogame.php , i transfer them to the game page. Now i want to be able to track each players progress via database so that when they log back in and they get to the transfertogame.php page, it can check the page they were on before and immediately take them to the page without starting from page 1. Any help will be appreciated as i am very poor with php scripting This is the session ID i guess is being used from my application page //Initialize Method @4-537EA73F function Initialize() { if(!$this->Visible) return; $this->ds->Parameters["sesUserID"] = CCGetSession("UserID"); } //End Initialize Method //Validate Method @4-7E1FC38C function Validate() { $Validation = true; $Where = ""; $this->CCSEventResult = CCGetEvent($this->CCSEvents, "OnValidate"); return (($this->Errors->Count() == 0) && $Validation); } //End Validate Method Hi, I want to be able to let user type in xml text and it will be parsed and uploaded to db, but it doesn't work, it just keeps redirecting me back to this form below! here is html form: Code: [Select] <html> <body> <form method='post' method='form.php'> <p> <textarea name="pastedXML" rows="10" cols="30"> Please paste your xml file here. </textarea> </p> <p> <input type="submit" value="Convert to SQL" name="textXML" /> </p> </form> </body> </html> Here is script (form.php;I just want to retrieve the contents typed in text area to store to variable...how??) Code: [Select] <?php //get the text in textarea and shred it! if(isset($_POST['textXML'])) print $_POST['pastedXML']; ?> Any help much appreciated! hey i am using a MySql database and i need to create a dynamic HTML table with one of its columns as checkboxes.so i have to create multiple checkboxes.but these checkbox values are to be stored in a mysql table and then later retrieved when form reloads.and depending on previous state when form was submitted, the newly created checkboxes have to be checked in the same manner.so how do i store multiple checkbox values in my table and also how do i retrieve them? please help. Hello everybody!
I am trying to figure out how to obtain all the data related to a key, but I've got no results so far and I am becoming really frustated, let's see if any of you could help me out with this.
Imagine I have a table with several columns, but we bother about two of them, let's say we have serial numbers of some product, on the left Incoming serial number (we can repair or swap the unit), on the right the outcoming serial number (same if we have repared the unit, different if we swap it for another unit).
Then we have, for example:
A -> A (Unit A enters and we repaired it)
A -> B (Same unit came another day for some reason and we couldn't repair it, so we swap it by giving B to the customer)
B -> C (Unluckily B was defective so we have to change it again)
C -> C (C had another problem and we repaired it)
We have that in the database from different days and the such, so now, we want to know the historical and we know "C". If we perform a SELECT * FROM... WHERE incoming/outcoming serial number = "C" we'll get:
B -> C
C -> C
So we should seek now for B and keep going... but I cannot proceed correctly, 'cause if I SELECT using B I'll get again B -> C (and A -> B, what I want), but when do I know I have to finish? How could I implement this as a function or whatever? showing every not repeated line from the beginning.
Could your minds help mine? Thank you very much in advance!.
I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks I want to use session to do a query and will I be able to do this? I have a session that was gathered from login and now i was to use this session to do a query If Yes, How? Hi folks, Complete No0b here when it comes to PHP and MySQL. I am in the middle of creating a PHP website. What I want to do is have the contents of a page in a MySQL table and have PHP gather the page content and display it on the page. Here is what I have in my home.tpl file: Code: [Select] <?php $query = "SELECT home, FROM $database_name"; $result = mysql_query($query); ?> And, in my index.php I simply call that home.tpl to display the data by using: Code: [Select] <?php include 'templates/default/home.tpl'; ?> Now, all I get is a blank page on index.php. Is there something else I should be doing? Remember, I am a complete No0b! Thanks folks. Hi everyone. I've two MySQL tables (tbl_csv_input & tbl_qn_types). Table Details: 1. tbl_csv_input (qid, qname, qn_answer, qn_type) Primary Key = qid; Foreign Key = qn_type ref tbl_qn_types qn_type_id 2. tbl_qn_types (qn_type_id, qn_type_name, notes) Primary Key = qn_type_id Sample Data: tbl_csv_input 1 ; Ferrari is the fastest car? ; Yes||No ; T1 2 ; Which below features would you like to have in you... ; Navigator||Airbag||Seat Belt||Camera & Sensors ; T1 3 ; Which model would you prefer? ; CX||MX||SX||LX ; T2 4 ; Comments ; ; T3 tbl_qn_types T1; checkbox; This qn type is used for yes/no T2; radio; Multiple options but only one is correct T3; text area; Users enter input like comments The whole idea is to have a questionnaire displayed in a HTML table depending upon the question types (T1, T2, T3) In simple terms: You have a question and below it there are options. Some questions have check boxes and some have radio buttons and some have text areas. The problem I'm facing is with the column: qn_answer; and column = qn_type. I'm unable to make a loop inside a table which is already in a loop. Please see the attached files (which has the code I've written). TIA. [attachment deleted by admin] Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago. I started out with some tutorials and feeling im getting the hang of it. Enough of me, lets get to the point: <?php $con = mysql_connect('localhost',$user,$pass)or die(mysql_error()); $selectdb = mysql_select_db($selectdb)or die(mysql_error()); $sql = "SELECT * From table"; $result = mysql_query($sql); $num = mysql_num_rows($result); $myarray = array($result); $i =0; while ($i < $num){ echo $myarray[$i]; $i++; } ?> Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results. The output in the browser is this: Resource id #3 I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple". Thanks in advance! Dan-Levi I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> I wonder whether someone can help me please. I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page. Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be: Code: [Select] <?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?> The problem I have, is that when I try and run this, I receive the following error: Quote Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";. I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem. I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong. Many thanks and kind regards Hi, how do I store as a string (in a variable) a mysql query b/c what I'm doing below outputs Resource id in client browser: Code: [Select] <?php //database connection set up etc $show=mysql_query("SELECT file_Name FROM xdocument WHERE doc_id=95"); print $show; ?> Any help much appreciated, thanks. hi i want to store url to images in database for logged in users (where id = $id) and recall the image hopefully using --------------------- <img src="<?php echo row['link']; ?>" /> or similar and need help with the sql update string any ideas please help i been stuck with this for some time and now decided to ask around in this forum for help, please help if you can. I am using wampserver 2.2 on window 7 64 bit machine. I am using jpegcam to take snapshots and it works fine. I make upload folder within jpegcam/htdocs/upload to store snapshots. All works fine but how can i store uploaded images into mysql. I also want to retrieve image from mysql and print in php page. The problem is jpegcam used java script in form in which POST method and type file was not used. I am already done task like upload images to folder and store it into mysql using POST an $_FILES varibales . here the code of in jpegcam/htocs/test.html <form name="f1" enctype="multipart/form-data" method="" action="" > <input type=button value="Configure..." onClick="webcam.configure()"> <!-- <input type=button name="img" value="Take Snapshot" onClick="take_snapshot()"> --> <a href="javascript:void(webcam.freeze())">Freeze</a> <a href="javascript:void(webcam.upload())">Upload</a> <a href="javascript:void(webcam.reset())">Reset</a> </form> and this the code of jpegcam/htocs/test.php $target = 'upload/'; $filename = $target . date('YmdHis') . '.jpg'; $result = file_put_contents( $filename, file_get_contents('php://input') ); if (!$result) { print "ERROR: Failed to write data to $filename, check permissions\n"; exit(); } Now please tell me how can i store snapshots in mysql an retrieve it an print it in another php page.please plzplz help me as soon as possibe. please please... Thanks in advance. I'm trying to do something that I thought was very simple about 2 weeks ago :-( I want to put a form on my site and link it to a database so when a user types a surname into the form they can search the db and the page will only display the surnames that match the search criteria. I got the db set up using phpmyadmin in about 2 minutes flat, but keep getting error messages when I write the php. I'm currently working with the below script, which keeps giving me the error 'unexpected T_string' on line 176 I've searched every forum and help site I can find, and the mysql and php manuals are just mind-boggling. I'm sure I'm making some really amateur mistake, but I'd really appreciate help with this! thanks all! <html> <head> <title>SEARCH RECORDS</title> </head> <body> <FORM NAME ="Search" METHOD ="POST" ACTION = "test3"> <INPUT TYPE = "TEXT" VALUE ="surname" NAME = "surname"> <INPUT TYPE = "Submit" Name = "Search" VALUE = "Search"> </FORM> </body> </html> <? $user_name = "*****"; $password = "*****"; $database = "*****"; $server = "localhost"; $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM *****" WHERE surname=$_POST['surname']; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { print $db_field['Grave'] . "<BR>"; print $db_field['Surname'] . "<BR>"; print $db_field['Forenames'] . "<BR>"; print $db_field['Death_Date'] . "<BR>"; print $db_field['Birth_Year'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?> |
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