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PHP - Execute Php Stored In Mysql Database?
Hi Guys,
I have a webpage which has subsequent pages stored in a database e.g. index.php?id=1 The problem being, is that id=1 has it's data pulled from a database. This was fine & dandy until I started to insert PHP...I am trying to get the below to executre <?php $rater_id=1; $rater_item_name='Rate This'; include("rater.php");?> However nothing shows & nothing happens, I know eval can be used but have not been succesfull in implementing this, can someone please help! Similar TutorialsHi! I was wondering if there is a way to execute php code which is stored in mysql database using php. At the minute I am using a echo to try and run php code stored in a mysql database but this just displays the code and does not run the php code. Thanks for any help! How can I run php code that is stored in a database? I have my pages created by grabbing the contents of a pageBody mySQL table cell, but some pages require further php to be able to display correctly. For example, I have a players page, which will list all players, and some information about them. The players and information is all saved in a players table in my database (separate to the pages table where pageBody is stored). I use simple php to grab all the players and format all of their information so it can be displayed on the page. This is the flow of information that I would like: User clicks on players page browser loads content page (this is a template page), and grabs the pageid from $_GET browser then reads pages table in database to get the pageBody associated with the pageid The pageBody will contain more php, which will run the players script to retrieve the list, and display Doing it the above way will make it much easier to extend the website to include more types of pages that has to run additional php scripts. The only way I can think of this working is by doing this: user clicks on players page browser loads content page, grabs pageid from $_GET browser checks the pageid for the one associated with players (hard coded into the php script) browser then loads the players.php script instead of going to the database This above way means that I will need to edit the index.php page everytime I add a new list type (example, coaches, volunteers etc). Anyone have any suggestions? I know my question is long, but I was finding it hard to explain it in an easy to understand way, but I'm still not sure if I have :S. Cheers Denno Hey guys, I have been banging my head against a wall here with this. I am saving my sessions in my database via session_set_save_handler(). So let me walk you through what I have here, what works, and what the issue seems to be. basically, the problem is the $_SESSION array is empty on page load. common.php: I have the open / close / read / write / destroy / and gc functions. These all work properly as when i use a session variable it stores into the database and i can see all of the information in there.. inside of common.php i have session_start().. I am positive that the session_start() is running becasue common.php is included into index.php and i tried adding session_start() to index.php again and i got an E_NOTICE saying the session already began. (yes, for the read function i am returning a string... i included that below). for the table itself, i have set the session_data as both text and blob, same issue with both. index.php: includes common.php. I know it's included as other aspects of the file are included and work properly. if i call var_dump($_SESSION) i get an empty array. and i prited out the session_id() and it matched my session id in the database. and the values that are stored in there are correct. i have for example: count|i:0 now with that value actually in the database and when calling session_id() and i get the ID that matches in the table. so i will get an E_NOTICE of an undefined index.. i was trying something like: if(!isset($_SESSION['count'])) $_SESSION['count'] = 0; else $_SESSION['count']++; echo $_SESSION['count']; Everytime the page is reloaded, count is reset to 0.. I can tell that it is reset to 0 as the expired time changes in the database after every load of the page (which is part of the write function). I double checked that it wasn't a problem for some reason with the ++ by adding in a variable that sets to 1 when in the if, and 0 if in the else, and it always outputs a 1. i have included here my read function since that apparently is the issue.. function sess_read($sess_id) { global $DB; $sql = "SELECT `session_data` FROM `sessions` WHERE session_id = '".$sess_id."' AND session_agent = '".$_SERVER["HTTP_USER_AGENT"]."' AND session_expire > '".time()."';"; $query = $DB->query($sql); if($DB->num_rows($query) > 0) { $r = $DB->fetch($query); return settype($r['session_data'], 'string'); } return ''; } I tried also with removing the agent and expire check to see if it was an issue there but same problem. I probably have missed something really dumb but i can't for the life of me figure it out and I have googled around for a similar issue. The actual output of the page is correct.. all of the HTML and CSS information loads properly.. no errors are reported (and i have E_ALL on). Thanks Hi Everyone, I have a program that generates 200 unique images keeping the first image static in each run.The images keep scrolling on to the screen pause for 3 seconds and scroll off I'm able to generated all 200 unique images without repetition, everything is working well except for the lase two images the last two images are scrolling on to the screen but are not been displayed in the database, Moreover The last image is a duplicate of 197th image.I don't know what is happening..... Here is MY code.......... <?php session_start(); $sid = $_SESSION['id']; $_SESSION['imageDispCnt'] = 0; $myQuery = "SELECT * from image"; $conn = mysql_connect("localhost","User","Passwd"); mysql_select_db("database_Name",$conn); $result = mysql_query($myQuery); $img =Array(); $id =Array(); $i =0; $imagepath = 'http://localhost/images/'; while ($row = mysql_fetch_array($result)) { $img[$i] = $imagepath.$row['img_name']; $id[$i] = $row['imageid']; $i = $i + 1; } ?> </head> <script language="JavaScript1.2"> var scrollerwidth='800px'; var scrollerheight='600px'; var scrollerbgcolor='white'; var pausebetweenimages=3200; var s; var sec; var d; var j; var imgid; var milisec = 0; var seconds = 0; var flag = 1; var ses_id = '<?php echo $sid;?>'; var count = 0; var i = 0; var imgname; var imgid; var k =0; var slideimages=new Array(); var img_id = new Array(); var index; <?php $l =0; $count = array(); $j = rand(0,199); while($l < 200) { while(in_array($j, $count)) { $j = rand(0,199); } $count[$l] = $j; $l++; }?> <?php $k = 0; for($k = 0;$k<count($count);$k++){ ?> index = <?php echo $k;?>; <?php $indx = $count[$k];?> if(index == 0){ slideimages[0] = '<img src="http://localhost/images/hbag044.jpg" name="r_img" id="0"/>'; img_id[0] = '<input type="hidden" value="0" id="imgId" />'; } else if(index > 0) { slideimages[<?php echo $k?>] = '<img src="<?php echo $img[$indx]?>" name="r_img" id="<?php echo $id[$indx]?>"/>'; img_id[<?php echo $k?>] = '<input type="hidden" value="<?php echo $id[$indx]?>" id="imgId" />'; } <?php } ?> Can Any one plese help me Appreciate your help... Thanks Hi, I am trying to store a sql query in a database, but every time I try to read it, it shows all the code as text. eg just print ' . $parts_row['part_number'] . ' Here is the code on the page:
$parts_sql = "SELECT * FROM parts WHERE part_cat = " . $category_row['cat_id'] . " ORDER BY CAST(part_a AS DECIMAL(10,2))";
$parts_result = mysql_query($parts_sql);
if(!$parts_result)
{
echo '<tr><td>The parts could not be displayed, please try again later.</tr></td></table>';
}
else
{
while($parts_row = mysql_fetch_array($parts_result))
{
$pageformat_sql = "SELECT * FROM category_pages WHERE catpage_number = " . $category_row['cat_page'] . "";
$pageformat_result = mysql_query($pageformat_sql);
if(!$pageformat_result)
{
echo 'Error';
}
else
{
while($pageformat_row = mysql_fetch_assoc($pageformat_result))
{
$data = $pageformat_row['catpage_table'];
echo '<table class="table2 centre" style="width:750px">
' . $pageformat_row['catpage_tabletitle'] . '' . $data . '';
}
}
}}
And this is what is stored in the database table:
<tr> <td class="cell left">' . $parts_row['part_number'] . '</td> <td class="cell centre">' . $parts_row['part_a'] . ' mm</td> <td class="cell centre">' . $parts_row['part_b'] . ' mm</td> <td class="cell centre">' . $parts_row['part_c'] . ' mm</td> <td class="cell centre">' . $parts_row['part_d'] . ' mm</td> <td class="cell centre">' . $parts_row['part_e'] . ' mm</td> <td class="cell centre">' . $parts_row['part_imped100'] . '</td> <td class="cell centre"><a href="datasheets/' . $parts_row['part_datasheet'] . '.pdf"><img border="0" src="images/pdf.jpg" alt="Download"</a></td></tr>I would be very grateful to anyone who can help me with this. I put together the following blocs of code for uploading pictures into a database and displaying them on a webpage. The pictures are supposed to be displayed on the member's only page of a website I'm working on, upon logging in, and they are supposed to be the member's uploaded picture. I created several members and and used one of my existing member accounts to test the uploading process. The picture upload process appeared to have been successful when I checked on myphpadmin. Yet, when I login with this account, no picture is displayed, instead, a tiny jpg icon is displayed at the top left corner of the box in which the picture was supposed to be displayed. Same thing when I login with the other accounts with which I haven't yet uploaded a picture. I'll start with the code that installs the table in the database $query = "CREATE TABLE images ( image_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY , member_id INT UNSIGNED, like_id INT UNSIGNED, image LONGBLOB NOT NULL, image_name varchar(255) NOT NULL, image_type varchar(4) NOT NULL, image_size int(8) NOT NULL, image_cartegory VARCHAR(20) NOT NULL, image_path VARCHAR(300), image_date DATE )"; Then here is the code which allows the member to upload his pictu <form enctype="multipart/form-data" action="insert_image.php" method="post" name="changer"> <input name="MAX_FILE_SIZE" value="102400" type="hidden"> <input name="image" accept="image/jpeg" type="file"> <input value="Submit" type="submit"> </form> And here is the insertimage.php which inserts the image into our database: Note that I have to authenticate the user in order to register his session id which is used later on in the select query to identify him and select the right image that corresponds to him. <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Make sure the user actually // selected and uploaded a file if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { // Temporary file name stored on the server $tmpName = $_FILES['image']['tmp_name']; // Read the file $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); // Create the query and insert // into our database. $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')"; $results = mysql_query($query); // Print results print "Thank you, your file has been uploaded."; } else { print "No image selected/uploaded"; } // Close our MySQL Link mysql_close(); } //End of if statmemnt. ?> On the page which is supposed to display the image upon login in, I inserted the following html code in the div that's supposed to contain the image: <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border- color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <img src=picscript.php?imname=potwoods> </div> And finally, the picscript.php contained the select query: <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require('config.php'); $image = stripslashes($_REQUEST[imname]); $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND cartegoty = 'main' "); $row = mysql_fetch_assoc($rs); $imagebytes = $row[image]; header("Content-type: image/jpeg"); print $imagebytes; ?> Now the million dollar question is, "What is preventing the picture from getting displayed?" I know this is a very lengthy and laborious problem to follow but I'm sure there is someone out there who can point out where I'm not getting it. Thanks. this is the line in my script that I have to show the image: Code: [Select] $output .= "<img>{$row['disp_pic']}</img></br>\n"; As you can see I added the image tag, but it wont show the actual image. IE shows it as a small square with another small square picture icon in the middle of it (i'm sure you guys know what i mean). i am developing online test ,after succesful completion of the test i want to display the correct answe,in databse i have stored corect answer as radio button values(ex 1,2,3,4,) i want to display the coreect answer with a right mark,how can i achive this?could anybody help?
here is my code
<?php $testid=$_GET['testid'];
include_once("header.php");
?>
<html>
<head>
<body>
<?php
include_once('connect.php');
$sqltest="select testname from test where testid='$testid'";
$sqltestresult=mysql_query($sqltest) or die(mysql_error());
while($result=mysql_fetch_array($sqltestresult))
{
$testname=$result['testname'];
}
?>
<!--pass test name and time left -->
<table width="100%" cellpadding="0" cellspacing="0" border="0"><tr><td width="50%" valign="middle">
<h1 style="margin-left:0;"><?php
if(isset($testname))
{
echo $testname ;
}
?></h1>
</td>
<td width="50%" align="right" valign="middle">
<ul class="tabs" id="tabs">
<li id="showtime" style="font-weight:bold; color:#FF0000; bottom:5px; font-size:16px; float:right"></li>
</ul>
</td>
</tr></table>
<div class="div-tabs-container" class="div-tabs-container" style="width:79%;margin-left:3%" >
<table cellpadding="0" cellspacing="0" border="0" id="questiontable" >
<tr>
<?php
include_once('connect.php');
$sql="select * from testquestions where testid='$testid'";
$result=mysql_query($sql) or die(mysql_error());
if (mysql_num_rows($result)>0)
{
$data = array(); // create a variable to hold the information
while (($row = mysql_fetch_array($result, MYSQL_ASSOC)) !== false)
{
$data[] = $row; // add the row in to the results (data) array
}
//shift off the first value
array_shift($data[0]);
$merge = call_user_func_array('array_merge', $data);
$size=sizeof($merge);
$newdata=array();
$num=1;
$correctanswerarray=array();
$selctedanswers=array();
/* echo "<form action='fetchquestion.php' method='post' id='formsubmit' onsubmit='return out()' name='test'>"; */
foreach ($merge as $key => $value )
{
$sql2="select qid, question,option1,option2,option3,option4,correct_answer from question where qid='$value' ";
$result2=mysql_query($sql2) or die(mysql_error());
while($row2=mysql_fetch_assoc($result2))
{
$questionid=$row2['qid'];
$question=$row2['question'];
$option1=$row2['option1'];
$option2=$row2['option2'];
$option3=$row2['option3'];
$option4=$row2['option4'];
$correctanswer=$row2['correct_answer'];
#array to store the correct answer.
$correctanswerarray[] = $questionid.$correctanswer;
#$newdata[]=$row2;
echo "
<table class='bix-tbl-container' cellspacing='0' cellpadding='0' border='0' width='100%' id='questiontable'><tr>
<td class='bix-td-qno jq-qno-2413' rowspan='2' valign='top' align='left'>$num</td>
<td class='bix-td-qtxt' valign='top'><p>$question</p></td>
</tr>
<tr>
<td class='bix-td-miscell' valign='top'><table class='bix-tbl-options' id='tblOption_2413' border='0' cellpadding='0' cellspacing='0' width='100%'><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_A_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='1' id='disable'/>
</td><td class='bix-td-option' width='1%'>A.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_A_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option1</td>
<td id='tdAnswerIMG_A_2413' class='jq-img-answer' valign='middle'style='display:none;padding-left:10px;'>
<img src='/_files/images/website/accept.png' alt='' />
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_B_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='2' id='disabble'/>
</td><td class='bix-td-option' width='1%'>B.</td>
<td class='bix-td-option' width='48%'id='tdOptionDt_B_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option2</td>
<td id='tdAnswerIMG_B_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt=''/>
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_C_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='3' id='disable'/>
</td><td class='bix-td-option' width='1%'>C.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_C_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option3</td>
<td id='tdAnswerIMG_C_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt='' />
</td>
</tr>
</table></td></tr><tr><td nowrap='nowrap' class='bix-td-option bix-td-radio' width='1%' id='tdOptionNo_D_2413'>
<input type='radio' class='result-option cls_2413' name='option_$questionid' value='4' id='disable'/>
</td><td class='bix-td-option' width='1%'>D.</td>
<td class='bix-td-option' width='48%' id='tdOptionDt_D_2413'><table border='0' cellpadding='0' cellspacing='0'>
<tr>
<td class='bix-inner-td-option'>$option4</td>
<td id='tdAnswerIMG_D_2413' class='jq-img-answer' valign='middle' style='display:none;padding-left:10px;'>
<img src='/_files/images/website/wrong.gif' alt='' />
</td>
</tr>
</table></td></tr></table>
\n
<input type='hidden' name='count' value='$num' id='count'>
<input type='hidden' name='queId[]' value='$questionid' id='queId'>
</td>
</tr>
</table>
";
$num++;
}
}
echo "
</tr>
</table>
";
#array of correct answrer
#echo "array of correct answer<br/>";
#print_r($correctanswerarray);
#echo "size of the coreect answer array";
$size1=sizeof($correctanswerarray);
echo "<br/>";
}
else
{
echo " no questions available for selected test";
}
?>
</div>
<!-- add div right side -->
<div style="height:500px;width:150px;float:right;margin-top:-500px;border:1px solid #99CCFF">
</div>
<div style="height:30px">
</div>
<!-- add div bottom -->
<div style="height:125px;border:1px solid #99CCFF">
</div>
<?php
include('footer.html');
?>
</body>
</html>
Edited by mac_gyver, 25 October 2014 - 10:05 AM. code tags around code please Good afternoon all!! I have a very specific issue that, in my opinion, is quite complicated. In words, here is what I wish to achieve. I would like a page which displays users from the database. Each individual user has their own Div hidden below their name. Within this div, there is a form. The form is full of radio buttons. Once the user's name is clicked and form submitted, I wish to write the form data to the database for the specific user that was selected. See below example for better understanding: Dale Gibbs Chris Hansen Steve Jobs If I click Chris Hansen, the following happens Dale Gibbs ============== Chris Hansen HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT HIDDEN DIV FORM CONTENT Submit Button ============== Steve Jobbs So as of now, the display is correct. I am seeing exactly what I want to see from my database. Also, my div IDs work just fine as well as the Javascript toggleSlidebox function. The only issue is for some reason, whenever I submit the form (no matter which user I select from my list), I can only write to the last inputted ID. So for example, if the last ID entered into the database was 6, then thats the only ID that will be returned when my form is submitted and the only place data will be written to, even if I select a user with ID 2. Please see below code for more information Code: [Select] <?php $staff_display_query = "SELECT staff_info.id, staff_info.fname, staff_info.lname FROM staff_info, staff_projects WHERE staff_info.id = staff_projects.staff_id AND staff_projects.proj_id = '$c_project_id'"; $staff_display_sql = mysql_query($staff_display_query) or die (mysql_error()); while ($row = mysql_fetch_array($staff_display_sql)) { $current_staff_id = $row['id']; $staff_fname = $row['fname']; $staff_lname = $row['lname']; $list_staff .= ' ' . $current_staff_id . '<br /> <a href="#" onclick="return false" onmousedown="javascript:toggleSlideBox(' . $current_staff_id . ');">' . $staff_fname . ' ' . $staff_lname . '</a> <div class="hiddenDiv" id="' . $current_staff_id . '" style="border:#FFF 2px solid; width:553px;"> <!--TASK 1--> <div id="task_1_permissions" class="task_permissions"> <input name="permissions_1" type="radio" value="1"/> <input name="permissions_1" type="radio" value="2" /> <input name="permissions_1" type="radio" value="3" /> <input name="permissions_1" type="radio" value="0" /> </div> <!--TASK 2--> <div id="task_2_permissions" class="task_permissions"> <input name="permissions_2" type="radio" value="1"/> <input name="permissions_2" type="radio" value="2" /> <input name="permissions_2" type="radio" value="3" /> <input name="permissions_2" type="radio" value="0" /> </div> <!--TASK 3--> <div id="task_3_permissions" class="task_permissions"> <input name="permissions_3" type="radio" value="1"/> <input name="permissions_3" type="radio" value="2" /> <input name="permissions_3" type="radio" value="3" /> <input name="permissions_3" type="radio" value="0" /> </div> <!--TASK 4--> <div id="task_4_permissions" class="task_permissions"> <input name="permissions_4" type="radio" value="1"/> <input name="permissions_4" type="radio" value="2" /> <input name="permissions_4" type="radio" value="3" /> <input name="permissions_4" type="radio" value="0" /> </div> <input name="submit_user_permissions" type="submit" value="Submit Permissions" /> </div> </div> <br /><br /> '; } if (isset($_POST['submit_user_permissions'])) { $permissions_1 = $_POST['permissions_1']; $permissions_2 = $_POST['permissions_2']; $permissions_3 = $_POST['permissions_3']; $permissions_4 = $_POST['permissions_4']; $query = "UPDATE staff_projects SET task_1='$permissions_1', task_2='$permissions_2', task_3='$permissions_3', task_4='$permissions_4' WHERE proj_id='$c_project_id' AND staff_id='$current_staff_id'"; $sql = mysql_query($query) or die (mysql_error()); echo 'Permissions set successfully.<br />'; } After this PHP I have my standard HTML. Below is the javascript function I use for the slide box: Code: [Select] <script src="js/jquery-1.5.js" type="text/javascript"></script> <script language="javascript" type="text/javascript"> function toggleSlideBox(x) { if ($('#'+x).is(":hidden")) { $(".hiddenDiv").slideUp(200); $('#'+x).slideDown(300); } else { $('#'+x).slideUp(300); } } </script> Javascript works just fine by the way. Below is the form I have in the HTML of the code. Code: [Select] <form action="" method="post" enctype="multipart/form-data"> <?php echo "$list_staff"; ?> </form> This is of course wrapped around body tags and all the other necessary HTML. The view of the form is working right, the functions within the form are also working correctly. I just cant seem to separate the variables for each individual user. If I am in Chris Hansen's div, it should be his ID number that is being referenced for the database, not the ID number of the last person entered into the system. Any ideas? As always, thanks in advance guys!!! Bl4ck Maj1k Hi all, I've got a website for an event, each team have their details on a page which are recalled from a SQl database. But I'm wanting to create a password input box for each team, so when they enter the correct password they are taken to a page containing forms where they can edit the team details. Here is the page with the users details on where they anter the password: http://www.wharncliffenetwork.co.uk/wrc/entered/team.php?id=8 I'm not sure how to code it, Can an IF statement be used? Anyone got any pointers? I'f been unsuccessful in finding a tutorial or something similar. Hope that makes sense :S Cheers. Hi guys, I have a file location stored in mysql. when i populate the table i need this file location to be a hyperlink to the file itself, so the visitors can click like a normal link and open the file in word and pdf (both formats stored). example of file location as in db "_private/Incident_Reports/Incident%20-%20Applecross%20-%2017%20December%202010%20-%20Website.doc" example of php code Code: [Select] echo $row['word_document']; any ideas would be really appreciated. sorry if i have posted this in the wrong place, wasnt sure wether to post here or mysql. I have made a php calendar, and i am now wanting it to show if there is an event on that day and if so show it in a tool tip. the tool tip is populated by what is in the title="" of the link so i need my events to be shown in there. I have figured out how to show the event but now i am stuck on how to show all events if there is more than one one that day, how i have set it seems to only the second event, probably as the second variable overwrote the first variable. this is the code i have at the minute .............. $result = mysql_query("SELECT * FROM events WHERE day='$day_num' AND month='$fullmonth' AND year='$year'") or die(mysql_error()); $rows= mysql_num_rows($result); if ($rows !="false"){ while($rowout = mysql_fetch_array($result)) { $todayis = $rowout['day'] ."-". $rowout['month'] ."-". $rowout['year'] ."<br>"; $title= $rowout['event'] ."<br><br>";} $firstl = "<a href='' title='". $todayis . $title ."'>"; $lastl = "</a>";} else {$firstl = ""; $lastl = "";} then to display the day and links ................. Code: [Select] <td><? echo $firstl; ?><? print $day_num; ?><? echo $lastl; ?></td> could some one be so kind and help me write it so that it displays all events? here is a link to the calendar, incase its needed. http://www.scripttesting.htmlstig.com/calendar/index.php Many thanks Carl Hello, First post here! I am using a script to store small files into a database. I know it is not highly recommended but this is the only solution I can find in this case. All seems fine when inserting but when retrieving all documents are corrupt. I am working with PHP 5 on Windows server. Part script after upload: (the document is uploaded to the server first and this is working fine) Code: [Select] if (!empty($_FILES)) { $folder = "kenya/docs_temp/";//including last / $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = dirname(__FILE__); $targetPath = substr($targetPath,0,strrpos($targetPath, "uploadify")). $folder; $targetFile = str_replace('\\','/',$targetPath) . $_FILES['Filedata']['name']; $fileName = $_FILES['Filedata']['name']; $tmpName = $_FILES['Filedata']['tmp_name']; $fileSize = $_FILES['Filedata']['size']; $fileType = $_FILES['Filedata']['type']; $conn=odbc_connect('xxx','xxx','xxxx'); $content = file_get_contents($tmpName,true); $content = addslashes($content); if(!get_magic_quotes_gpc()){ $fileName = mysql_real_escape_string($fileName); } $file_info = pathinfo($_FILES['Filedata']['name']); $addDoc = "INSERT INTO tblfiles (file_name,file_type,file_size,file_content,file_extension) VALUES ('".addslashes($fileName)."','".$fileType."','".$fileSize."','".$content."','".$file_info['extension']."')"; odbc_exec($conn,$addDoc); } The the script to retrieve and output to the browser: Code: [Select] $conn=odbc_connect('xxxx','xxx','xxxx'); //get the document $getF = "SELECT * FROM tblfiles WHERE userId = '".$_GET['userId']."' AND fileId = '".$_GET['fileId']."'"; $rsF = odbc_exec($conn,$getF); $size = floor(odbc_result($rsF,'file_size')); $type = odbc_result($rsF,'file_type'); $name = odbc_result($rsF,'file_name'); $content = odbc_result($rsF,'file_content'); ob_end_clean(); header("Content-length: ".$size.""); header("Content-type: ".$type.""); header('Content-Disposition: attachment; filename="'.$name.'"'); echo $content; And finally here is the structure for the table: Code: [Select] CREATE TABLE `tblfiles` ( `fileId` int(11) NOT NULL auto_increment, `userId` int(11) NOT NULL, `file_name` varchar(100) NOT NULL, `file_type` varchar(100) default NULL, `file_size` int(11) default NULL, `file_content` mediumblob, `file_extension` varchar(50) NOT NULL, PRIMARY KEY (`fileId`) ) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ; many thanks for any assistance. I'll walk you first through what is working: <?php echo($user['biz_site'] != '' ? "<h4>Web Site</h4>" . $user['biz_site'] : ""); ?> In a results page, the code renders something that looks like: Quote Web Site http://www.somecoolsite.com What I'd like instead is something like this: Quote Web Site Visit Site ...where the "visit site" links to the site. I'm trying to address a problem of some sites, especially government sites, having 75+ character-long addresses, which messes up the finite space I have allotted for site addresses. Everything I've tried thus far has rendered blank white pages. (the original code posted up top does not produce a hyperlink) Okay, so here is the deal. Have a table which stores image as blob files. Now i want to read the image width and height directly from the blob field. Is this possible and if yes, how? Things i tried so far; list($size[0],$size[1],$type, $attr) = getimagesize('image.php?i=26ddd45b02859e836d13d4b9fde34281'); print_r($size); $img = 'image.php?i=26ddd45b02859e836d13d4b9fde34281'; echo imagesy($img); image.php grabs the image from DB and show's it with header("Content-type: image/jpg"); It works for just showing the images with the <img> tag. Any ideas of help would be great! Hi Please take a look at my code snippet. I don't know how to get the number of rows returned from a MySQL stored procedure so that I can handle it accordingly. Any ideas please? Thanks Let's say I have a mysql table named 'function' and a field named 'body' with 1 row. The 'body' field contains 'Echo out the number 1000 with a comma. ".number_format(1000).". Simple Example.' As you can see there is a function; number_format(); in the database, but is it possible to execute it onto a live webpage and be displayed correctly? Hi all: I have been going nuts trying to get this to work. I have gone through many attempts based on internet searches and examples that simply do not work. I have files stored in my MySQL db. The files cna be of any type and are being displayed in a listed report. The files are hyperlinked and clickable, but here is the issue: When I click on a file in: http://development.products-and-services.ca/cms/test.php I get: http://development.products-and-services.ca/cms/test2.php?content=DHall_CV_webdev_eng.pdf as a blank page result. The objective is to get a file download alert or to see the file in the browser. Any ideas? Hi All, I need to subtract dates and display the number of days left. I have a 'Start' date and an 'End' date in DATETIME format in the DB. Not quite sure where to start. A simply start - end doesn't work . Start = 2011-11-01-00:00:00 End = 2011-11-30-23:59:59 Since it is now 2011-11-27, my output should equal 3. Any help is appreciated. |
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