PHP - No Values Shown In <select> Aft Using Php Code
Hi all,
Currently my dob_day is not showing any value, even though there is a value in the database, it seems like it is unable to retrieve the value. No value was shown in the <select> box. Do you guys have any idea? Thanks Code: [Select] <?php $query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'"; $data = mysqli_query($dbc, $query) or die(mysqli_error($dbc)); // The user row was found so display the user data if (mysqli_num_rows($data) == 1) { $row = mysqli_fetch_array($data); if ($row != NULL) { $name = $row['name']; $nric = $row['nric']; $gender = $row['gender']; $old_picture = $row['picture']; $dob_day = $row['dob_day']; } else { echo '<p class="error">There was a problem accessing your profile.</p>'; } //HTML CODING <label for="dob" class="label">Date Of Birth</label> <select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>"> <option> <option>01</option> <option>02</option> <option>03</option> <option>04</option> <option>05</option> <option>06</option> <option>07</option> <option>08</option> <option>09</option> <option>10</option> <option>11</option> <option>12</option> <option>13</option> <option>14</option> <option>15</option> <option>16</option> <option>17</option> <option>18</option> <option>19</option> <option>20</option> <option>21</option> <option>22</option> <option>23</option> <option>24</option> <option>25</option> <option>26</option> <option>27</option> <option>28</option> <option>29</option> <option>30</option> <option>31</option> </select> ?> Similar TutorialsI need a little help here. I have a function currently that allows me to send data to a php script from my C++ application. How can I do the opposite? C++ Code below to send data to the php script: Code: [Select] void SendData() { char* Navigate = new char[500]; memset(( void* )Navigate, 0, 500); char* Version = "Version 1.0"; sprintf(Navigate, "MYSITE.COM/GetData.php?v=%s", Version); SendFunction(Navigate); } DWORD SendFunction(char* link) { char* szURL = link; HINTERNET hSession; HINTERNET hRequest; hSession = InternetOpen( // Make internet connection. "Microsoft Internet Explorer", // agent INTERNET_OPEN_TYPE_PRECONFIG, // access NULL, NULL, 0); // defaults if(hSession) { hRequest = InternetOpenUrl( // Make connection to desired page. hSession, // session handle link, // URL to access NULL, 0, 0, 0); // defaults } if(hRequest) { bool bResult = HttpSendRequest(hRequest, NULL, 0, NULL, 0); } InternetCloseHandle(hSession); return 0; } php script that receives the data from the C++ application: <? $f=fopen('FILE','a'); fwrite($f,$_GET['v'] \n); fclose($f); ?> This just simply sends the char* Version = "Version 1.0" text to the php script witch _GETs it and writes it to a file. Simple and impractical, however I just wanted to get something rolling. It works fine, now I need to do the opposite. I need to do exactly what this function does, but flip it around!. I want store a value in the php script, and get that value with my C++ application, and then log it out. Any help here please!? php script for what I want to do: <?php $dbhost = 'MyHost'; $dbuser = 'MyUserName'; $dbpass = 'MyPassword'; mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'MyDB'; mysql_select_db($dbname); $query = mysql_query("SELECT * FROM AppVersion"); while($row = mysql_fetch_array($query)) { $version = $row['Version Info']; } ?> I need to query the php script from my C++ application and retrieve the value of $version. How can I setup the php script to export that value $version for the C++ app and use the WinInet funcs to retrieve it? Note: I am just trying to get a feel for how all of this works right now, so none of this really has a practical use at the moment, but is helping me progress in my overall goal, so it will be very helpful to get this worked out. Hi all i have 2 small questions and maybe someone could help me out. Say I have a database with a table 'users'. This table has fields: email, name and year. What I want to do is show at the front-end all users, by selecting them from the table, and arrange them by year in column 1 and show them as individuals in column 2. But here is the thing; if someone selects for instance year 1970 and also someone from 1970 individually, someone will be selected twice. (edit:As a side note the selecting is aimed and emailing the users, that's why selecting someone twice is not good First question is: does anyone know how to remove doubles (maybe even triples if there is a select all button, maybe turn this around and say select only 1 instance of a user). Second question: is this a wise way to let someone select. Right now all users are fetched, and if the table users grows this could be tricky because of the large number of users. If someone could point me in the right direction of thinking I would be very happy, been thinking about this and scalability for quite a while now I have a Prepared Statement that runs a SELECT statement and returns 2 records, and I would like to store the Field Value for each Record into an Array. Here is the code I usually use for queries that return just a single value... // ****************** // Populate Form. * // ****************** // Build query. $q2 = "SELECT response FROM bio_answer WHERE member_id=?"; // Prepare statement. $stmt2 = mysqli_prepare($dbc, $q2); // Bind variable to query. mysqli_stmt_bind_param($stmt2, 'i', $memberID); // Execute query. mysqli_stmt_execute($stmt2); // Store results. mysqli_stmt_store_result($stmt2); // Check # of Records Returned. if (mysqli_stmt_num_rows($stmt2)>0){ // Details Found. // Bind result-set to variable. mysqli_stmt_bind_result($stmt2, $response); // Fetch record. mysqli_stmt_fetch($stmt2); // Close prepared statement. mysqli_stmt_close($stmt2); }else{ // Details Not Found. $_SESSION['resultsCode'] = 'DETAILS_NOT_FOUND_2133'; // Close prepared statement. mysqli_stmt_close($stmt2); // Set Error Source. $_SESSION['errorPage'] = $_SERVER['SCRIPT_NAME']; // Redirect to Display Outcome. header("Location: " . BASE_URL . "/members/results.php"); // End script. exit(); }//End of POPULATE FORM Can someone help me out with the syntax so that I get an end result like this... $answerArray[0] = 'I want to be my own boss!!' $answerArray[1] = 'Don't waste your time trying to do your own Taxes!' Thanks, Debbie Hi guys, I am trying to create a script that gets a multiple select form selections and assign them to variables with php via the POST method. This is my code: Code: [Select] <select name="search_commodity[]" multiple="multiple" size="3" style="position:absolute; top:80px; left:414px; " > <option value="all">All</option> <option value="none">None</option> <option value="some">Some</option> </select> Then after the form gets submited then it should go to this php page and the first value of the chosen options has to be assigned to the variable called commo. The problem is that is is not assigning anything. Code: [Select] <?php $commo = $_POST['search_commodity[0]']; ?> Can someone help please? I can't figure it out. hi people can vote 3 times with 3 select menus but i want them not to be able to choose 3x the same select option the options in the select menus come out of the database this is how i let them vote Code: [Select] <?php include("./includes/egl_inc.php"); $secure = new secure(); $secure->secureGlobals(); session_start(); if (isset ($_GET['match'])) { $matchid = (int)$_GET['match']; $_SESSION['matchid'] = $matchid; $matchmaps = mysql_fetch_array(mysql_query("SELECT maps,game FROM ffa_matches WHERE id=$matchid")); $matchmapschecker = $matchmaps[maps]; $matchgamechecker = $matchmaps[game]; $maps=mysql_query("SELECT id,map FROM ffa_maplists where gameid = $matchgamechecker"); while(list($id,$map)=mysql_fetch_row($maps)) { $maplist.="<option value='$id'>$map</option>\n"; } $out[body].="<table width='98%' border='0' cellspacing='0' cellpadding='0' style='border:1px solid black'> <tr style='color:#cccccc; background-color:black;'> <td valign='middle' background='$config[bg2]' align='center' colspan='10'> <br/>Choose Maps<br/> </td> </tr></table><table width='98%' border='0' cellspacing='0' cellpadding='0' style='border:1px solid black'> <tr style='background-color:black;'> <td> <form action='ffainsertmapvote.php' method='post'> <select name='map1'>$maplist</select> <select name='map2'>$maplist</select> <select name='map3'>$maplist</select> <input type='hidden' name='match' value='$matchid' /> <input type='submit' value='Vote' /> </form> </td> </tr></table>"; } include("$config[html]"); ?> do i need jquery to fix this? i dont want the user to be able to go to a next screen without having max 2 the same options selected thanks in advance ! Hi guys, Im developing a stock system and so far new stock and can be added with the quantity, however im trying to update the stock by using a select box and selecting the type of stock and inputting a new qty. the form and everything is set up and i can select files from the database, however i dont know how to update files from a select box The code i got so far is Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>stock manager</title> </head> <body> <?php $stockqty =&$_POST['stock_qty1']; ?> <center> <table> <td> <table> <td> <form action='stockview.php' method='POST'> Please Enter a Stock Name and Stock Value <table> <tr> <td> Stock Name: </td> <td> <input name="stock_name" type="text" /><BR /> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stock_qty" type="text" /> </td> </tr> </table> <input name="submit" type="submit" value="Add New Stock Items" /> </form> </td> </table> </td> <td> <table> <td> <form action='stockmanager.php' method='POST' enctype="multipart/form-data"> Please Select from the list the item you wish to update <table> <tr> <td> Stock Name: </td> <td> <?php $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query=("SELECT id, stockname FROM stocks"); $result = mysql_query ($query); echo "<select name=stock value=''>Edit Stock QTY</option>"; while($nt=mysql_fetch_array($result)) { //Array or records stored in $nt echo "<option value=$nt[id]>$nt[stockname]</option>"; /* Option values are added by looping through the array */ } $queryreg = mysql_query("SELECT * FROM stocks WHERE stockqty='$stockqty'"); $numrows = mysql_num_rows($queryreg); $update = mysql_query("UPDATE stocks SET stockqty='$stockqty' WHERE stockname = '$stockname'"); echo("Its updated"); ?> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stock_qty1" type="text" /> </td> </tr> </table> <input name="submit" type="submit" value="Update stock items" /> </form> </td> </table> </td> </table> </center> </body> </html> Any help would be greatly appreciated. Thanks Lance hi all please help... i would like to get datas between certain intervals. please have a look @ db_table.jpg, it's the db structure. the red highlighted part is the value i need. condition is select values when both device one & device two are off. db_table_report.jpg is the resulting report format. It contains start time & end time and the duration.. please help, i know it's a bit complicated, but it's urgent. I have set up a form page with a select box of colleges to select. I want the "options" in the select box to be values taken from a field called "name" in a table called "colleges" and they should be ordered alphabetically. I also want the default selected option to be "none." I have attached a picture to describe what i want. Please be detailed with the code. I am fairly new to php and mysql. Thank you. PHPBB3 does it. I crafted it on my own forum too. Why not have it here? Code is being posted daily. https://www.phpbb.co...655675#p5655675 See it in action? That would be usefull here instead of dragging and highlighting our mouse over the code and pressing CTRL+C. If IPB has a custom BBCODE manager, you could just use the code in their topic @Phillip. Edited by Monkuar, 25 January 2015 - 12:19 PM.
Example Code: if FirstImage=1 is absent, continue with FirstImage=0 Thanks for your idea and effort. please can you help .. I'm passing a value from another page using $_SESSION and trying to then use it in a SELECT statement in SQL think I'm missing something ... if hard code the value it works , and I have also checked at the other end that the variable is being assigned ...
code is (its the = $_SESSION['g_district'] "); that's causing the issue
I'm getting Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in /Applications/XAMPP/xamppfiles/htdocs/mytimekeeper/add/add.php on line 170 in the browser <div class="row s2 m2 l2"> <div> <label for="course_id">Choose the course raced:</label> <select name="course_id"> <?php // query to create course_id dropdown $resultset = $conn->query("SELECT `course_id`,`course_name` FROM `tbl_courses` WHERE `district`= $_SESSION['g_district'] "); while($row = $resultset->fetch_assoc()) { echo "<option value='{$row[course_id]}'>{$row['course_name']}</option>"; } ?> </select>
Hi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Hi,
I'm really at a loss here.
I have queried a table to get an option list, which returns what I expect but I need to then add it to some current code instead of the fixed option list presented.
The syntax is beyond me.
Please see attached.
Any pointers would be great.
Attached Files
OptionList.txt 1.61KB
7 downloads Howdy everyone, please i need help changing a php coded form from a checkbox to a select menu. Here's the form. <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" > <table class="dtable2"> <tr><th colspan="5">Enter a domain name:</th></tr> <tr><td colspan="5"><center>www.<input name="domain" type="text" size="35" /></center></td></tr> <tr><th colspan="5">Select an extension:</th></tr> <tr> <?php $i = 0; foreach ($this->serverList as $value) { if ($value['check'] == true) $checked=" checked "; else $checked = " "; echo '<td><input type="checkbox" name="top_'.$value['top'].'"'.$checked.'/>.'.$value['top'].'</td>'; $i++; if ($i > 4) { $i = 0; echo '</tr><tr>'; } } ?> </tr> </table> <center><input type="submit" name="submitBtn" class="sbtn" value="Check" /></center> </form> <?php I'll really appreciate your help. onlinegamekey. com/MTGT-Auction.php is the page I'm working on. The problem I'm having is cards with an apostrophe in the name breaks the operation. I am populating the Select Box with the Card Names and those are coming in fine, its not until I try to use the select value to get that specific card data do I have an issue. This query specifically Code: [Select] $quer2=mysql_query("SELECT * FROM auctions WHERE Card_Name ='$cards' Order By Price_Per") or die; I've tried $quer2=mysql_query("SELECT * FROM auctions WHERE Card_Name =" . htmlspecialchars($cards) . " Order By Price_Per") or die; but then I get no data for any card. Here is the page code I'm working with. Code: [Select] <?php $cards = $_POST['cards']; //SELECTING DATA FOR THE DROPDOWN $sql = "Select Card_Name From auctions Group BY Card_Name ASC" or die; $result = mysql_query($sql); ?> <script type="text/javascript"> <!-- var optList; var optsValue = new Array(); var optsText = new Array(); //when the page loads get the original options values and text and store them in arrays window.onload = function() { optList = document.getElementsByTagName("option"); for(var i=0; i<optList.length; i++) { optsValue[i] = optList[i].value; optsText[i] = optList[i].text.toLowerCase(); } } function searchSel(txtSearch) { //clear all the current options document.getElementById("items").options.length = 0; var count = 0; for(var i=0; i < optsValue.length; i=i+1) { if(optsText[i].indexOf(txtSearch.toLowerCase()) == 0) { //match found //add this option to the select list options var newOpt = new Option(optsValue[i],optsText[i],false,false); document.getElementById("items").options[count] = newOpt; count = count+1; } } } function reload(form) { var f1 = document.forms['f1'] var val=f1.cards.options[f1.cards.options.selectedIndex].value; self.location='MTGT-Auction.php?card=' + val ; } //--> </script> <style type="text/css"> body { background-color:#000000; } .row-one { background-color: #666666; font-family: Arial, Helvetica, sans-serif; font-size:12px; font-weight: bold; line-height: 17px; color:#CCFF33; } .row-two { background-color: #333333; font-family: Arial, Helvetica, sans-serif; font-size:12px; font-weight: bold; line-height: 17px; color: #FF0; } .th { background-color:#000000; font-family:Arial, Helvetica, sans-serif; font-size:14px; font-weight:bold; color:#CC0000; padding: 2; } </style> <!-- CREATE FORM & SELECT BOX --> <form method="post" name="f1" action="MTGT-Auction.php"> <select name="cards" id="items"> <option value='0'>Select...</option> <?php while ($row=mysql_fetch_array($result)) { if ($row['Card_Name']==@$cards) { echo "<option selected value='$row[Card_Name]'>$row[Card_Name]</option>"; } else { echo "<option value='$row[Card_Name]'>$row[Card_Name]</option>"; } } ?> </select> <br /> <input type="text" id="txt" value="Card Name?" onfocus="this.value==this.defaultValue?this.value='' :null" onkeyup="searchSel(this.value);" style="color:#000000; font:Arial; font-size:12px; background-color:#e1e1e1;" /> <BR /> <input type="submit" value="Submit" name="submit" /> <input type=button onClick="location.href='MTGT-Auction.php'" value='Reset' /> </form> <!-- CREATE TABLE WHERE DATA GOES --> <table border="1" bordercolor="#000000"> <tr align="center"> <th class="th">Auction ID</th> <th class="th">Card Name</th> <th class="th">Cards Per Auction</th> <th class="th">Auction Price</th> <th class="th">Cost Per Card</th> <th class="th">Date Listed</th> <th class="th">Seller Name</th> </tr> <?php //GET DATA FOR TABLE BASED ON SELECTED CARD & LOOP THROUGH $quer2=mysql_query("SELECT * FROM auctions WHERE Card_Name ='$cards' Order By Price_Per") or die; $i =1; WHILE($row = mysql_fetch_array($quer2)) { if ($i%2 !=0) $rowColor = "class='row-one'"; else $rowColor = "class='row-two'"; echo "<tr $rowColor>" . "<td>" . $row[Auction_ID] . "</td><td>" . $row[Card_Name] . "</td><td>" . $row[Qty_Listed] . "</td><td>" . $row[Price] . "</td><td>" . $row[Price_Per] . "</td><td>" . $row[Date] . "</td><td>" . $row[Seller] . "</td></tr>"; $i++; } //} ?> <?php //QUICK CHECK IS OUR VARIABLE SET??? echo "<font color=\"#FFFFFF\">". $cards . "</font>"; ?> </table> I image this is probably a very common problem & easy fix that has been answered many times, but I haven't found any thing that worked for me so any help.. or links to similar issues would really be appreciated. Thank you, I have a sql query similar to below in a .php file.
$sql = "select id,organisation,price from table where category = '$category';
These are the four unique values of the category column for your reference.
"A-t1"
"B-t1"
"C-t1"
"D-t1"
Now I want to create a dropdown list or listbox in .php file with option to select multiple values. If customer selects multiple values, it should fetch query for the selected categories.
E.g. dropdown list should be similar to below :
"A-t1"
"B-t1"
"C-t1"
"D-t1"
If an user selects "A-t1" and "C-t1", it should give output/query for the selected categories.
Here is the User profile search form... when i search that all details comes without Employee's Photo.. What's the error ? Code: [Select] </head> <body> <?php $uname = $_POST['uname']; $sql = "SELECT * FROM emp WHERE uname = '$uname'"; $result = mysql_query($sql); if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_array($result)) { $fname = $row['fname']; $lname = $row['lname']; $email = $row['email']; $address = $row['address_1']; $address2= $row['address_2']; $address3= $row['address_3']; $image = $row['image']; $gender=$row['gender']; $mobile=$row['tp_mobile']; $land=$row['tp_lnd']; $bod=$row['bod']; } } ?> <div class="main"> <div class="header"> <div class="header_resize"> <div class="logo"><h1></h1></div> <div class="clr"></div> <div class="search"> <form id="form1" name="form1" method="post" action="emp_profile.php"> <span> <input name="uname" type="text" class="keywords" id="textfield" maxlength="50" /> </span> <input name="image" type="image" src="images/search.gif" class="button" /> </form> </div> <div class="menu"> <ul> <!-- <li><a href="index.html" class="active">Home</a></li> <li><a href="services.html">Services</a></li> <li><a href="about.html"> About Us </a></li> <li><a href="contact.html"> Contact Us</a></li> --> </ul> </div> </div> <div class="clr"></div> </div> <div class="clr"></div> <div class="body"> <div class="body_resize"> <form id="" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype="multipart/form-data" accept-charset='UTF-8'> <div id="container"> <fieldset > <div id="content-container1"> <div id="content-container2"> <div id="section-navigation"> <img align="top" width="175" height="200" src="emp_images/<?php echo $image; ?>"/> <div> <!-- <label>Patient Name : </label> --> <br /> </div> </div> <div id="content" align="justify"> <h2> <label for="fname"> <?php echo $fname; ?></label> <label for="lname"> <?php echo $lname; ?></label> </h2> <p> <label><b>Email:</b></label> <br /> <label for="email"> <?php echo $email; ?></label> </p> <p> <label><b>Employee Address:</b></label> <br /> <label > <?php echo $address; ?></label> </p> <p> <label><b>Employee Address 2:</b></label> <br /> <label > <?php echo $address2; ?></label> <p> <label><b>Employee Address 3:</b></label> <br /> <label > <?php echo $address3; ?></label> <p> <label><b>Gender:</b></label> <br /> <label > <?php echo $gender ?></label> </p> <p> <label><b>Employees Mobile:</b></label> <br /> <label > <?php echo $mobile ?></label> </p> <p> <label><b>Employees Resident Number:</b></label> <br /> <label > <?php echo $land ?></label> </p> <p> <label><b>Date of Birth:</b></label> <br /> <label > <?php echo $bod ?></label> </p> </div> </div> </fieldset> <div class="clr"></div> </div> </div> </form> <div class="footer"> <div class="footer_resize"> <p>© P2011-001</p> </div> <div class="clr"></div> </div> </div> </div> </div> </body> </html> <?php mysql_close($con); ?> hi, i have the code below where it shows the images in connection with a reference. in my img table i only have two rows but the code below keeps showing it 7 times can you please help me? <table width=200> <div id="wrapper"> <!-- Content area --> <div id="content"> <!-- We will be working inside these tags --> <div id="gallery"> <?php $images=mysql_query("SELECT img.id, img.refimage, img.image, img.thumb, users.users_id FROM img, users WHERE img.refimage='$ref' AND users_id='$user_id'"); echo '<table width="40">'; while($row = mysql_fetch_array($images)) { $image=$row['image']; $thumb=$row['thumb']; echo '<div class="image_container">'; echo "<a href='$image' rel='lightbox[roadtrip]'><img src= '$thumb' width='60' height='40' alt='$title'>"; echo '</a></div>'; } ?> </div> <!-- We will be working inside these tags --> </div> </div> </table></td> Hello all , I am facing a problem when trying to to show image files generated from php code in IE8 . it is working fine on firefox and IE9 . but in IE8 its showing the famous red X instead of the image ; belw is the code: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html dir="RTL" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:v="urn:schemas-microsoft-com:vml" lang="en-us" > <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/> <head> <title>header </title> <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <?php mb_internal_encoding('UTF-8'); /** * @author IT Department * @copyright 2011 */ INCLUDE ("header.php"); INCLUDE ("connect.php"); if (isset($_GET['id'])) { $ID=$_GET['id']; $sql="select employees.USER_ID, employees.NAME_ARAB, employees.NAME_ENG, employees.TITLE_ARAB, employees.TITLE_ENG, department.DEP_NAME_ENG, department.DEP_NAME_ARAB, location.LOC_DESC_ENG, location.LOC_DESC_ARAB, employees.MOBILE, employees.EMAIL, employees.DEP, employees.LOC FROM employees,location,department WHERE employees.DEP=department.DEP_ID AND employees.LOC=location.LOC_ID and employees.USER_ID=$ID "; //echo $sql; $result=mysql_query($sql,$link); mysql_query("SET NAMES 'utf8'"); ECHO '<html DIR="RTL">'; echo "<table width=100% border='1' cellpadding='8' align='center' bgcolor='white'>"; while ($row = mysql_fetch_array($result, MYSQL_NUM)) { //$row = mysql_fetch_assoc($result); echo"<tr align='center'>"; echo"<td>"; ECHO " الرقم الوظيفي :"; echo"</td>"; echo"<td colspan='2'>"; ECHO "<strong>$row[0]</strong>"; echo"</td>"; echo"<td>"; ECHO " : User ID "; echo"</td>"; echo"<td rowspan='7'>"; ECHO "<img src='IMAGE.php?id=".$row[0]."' width='225' height='300'/>"; //the image echo"</td>"; echo "</tr>"; the php image file is : Code: [Select] <?php INCLUDE ("connect.php"); $ID = "'".$_GET['id']."'"; $sql="SELECT PHOTO FROM employees WHERE USER_ID=$ID"; $result=mysql_query($sql,$link); $row = mysql_fetch_array($result); header("Content-type: image/pjpeg"); print ( $row[0]); ?> Thanks in advance for your help So. I'm using this code right he <?php $imagepath = "css/images/statspage.jpg"; $image = imagecreatefromjpeg($imagepath); $imgheight = imagesy($image); $color = imagecolorallocate($image, 255, 255, 255); imagestring($image, 5, 70, $imgheight-50, "This is a test", $color); header('Content-Type: image/jpeg'); imagejpeg($image); ?> It works like it should. Now my problem is, is that the image is being displayed at a very low quality! Here's the original image: Here's the image with the code above: RIGHT away you can tell the HUGE difference. I'm not sure what I'm doing wrong. Maybe i need a higher res image? If y'all guys know a different method THAT also achieves the same thing I'm trying to do. Please let me know ASAP! |