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PHP - Display Selected Fields From Multiple Mysql Table
hi, i'm new to php. i've got a problem.
I've 3 tables and table has following content table 1 id name email address phone execution_date executor_name web_address table 2 id name email address phone execution_date executor_name project_title table 3 id name email address phone execution_date executor_name reviewer I want to export these table to and spreadsheet (.xls) with some criteria a form will assign which data will published at the spreadsheet with these criteria #all data can be exported from three tables #some column can be selected from three tables #some or all data can be selected from individual table i've implement a script for this but it didn't meet my requirements. Can anyone help? here is my script <?php $DB_Server = "localhost"; //your MySQL Server $DB_Username = "root"; //your MySQL User Name $DB_Password = "pass"; //your MySQL Password $DB_DBName = "mydb"; //your MySQL Database Name $search_from_date = $_POST['start_date']; $search_to_date = $_POST['end_date']; $Connect = @mysql_connect($DB_Server, $DB_Username, $DB_Password) or die("Couldn't connect to MySQL:<br>" . mysql_error() . "<br>" . mysql_errno()); $Db = @mysql_select_db($DB_DBName, $Connect) or die("Couldn't select database:<br>" . mysql_error(). "<br>" . mysql_errno()); $now_date = date('m-d-Y H:i'); if(($_POST['typer_of_report'] == 'rrc_report') || ($_POST['typer_of_report'] == 'all_report')) { $DB_TBLName = 'rrc_record'; } if(($_POST['typer_of_report'] == 'erc_report') || ($_POST['typer_of_report'] == 'all_report')) { $DB_TBLName2 = 'erc_record'; } if(($_POST['typer_of_report'] == 'aeec_report') || ($_POST['typer_of_report'] == 'all_report')) { $DB_TBLName3 = 'aeec_record'; } $file_type = "vnd.ms-excel"; $file_ending = "xls"; //} header("Content-Type: application/$file_type"); header("Content-Disposition: attachment; filename=protocol_report.$file_ending"); header("Pragma: no-cache"); header("Expires: 0"); if($DB_TBLName) { $sql = "Select * from ".$DB_TBLName." where execution_date >= '".$search_from_date."' and execution_date <= '".$search_to_date."' order by execution_date desc"; $Use_Title = 1; $title = "Report for $DB_TBLName on $now_date"; $result = @mysql_query($sql,$Connect) or die("Couldn't execute query:<br>" . mysql_error(). "<br>" . mysql_errno()); if ($Use_Title == 1) { echo("$title\n"); } $sep = "\t"; //tabbed character for ($i = 0; $i < mysql_num_fields($result); $i++) { echo mysql_field_name($result,$i) . "\t"; } print("\n"); while($row = mysql_fetch_row($result)) { $schema_insert = ""; for($j=0; $j<mysql_num_fields($result);$j++) { if(!isset($row[$j])) $schema_insert .= "NULL".$sep; elseif ($row[$j] != "") $schema_insert .= "$row[$j]".$sep; else $schema_insert .= "".$sep; } $schema_insert = str_replace($sep."$", "", $schema_insert); $schema_insert = preg_replace("/\r\n|\n\r|\n|\r/", " ", $schema_insert); $schema_insert .= "\t"; print(trim($schema_insert)); print "\n"; } //} echo "\n...\n"; } if($DB_TBLName2) { $sql = "Select * from ".$DB_TBLName2." where execution_date >= '".$search_from_date."' and execution_date <= '".$search_to_date."' order by execution_date desc"; $result = @mysql_query($sql,$Connect) or die("Couldn't execute query:<br>" . mysql_error(). "<br>" . mysql_errno()); $Use_Title = 1; $title = "Report for $DB_TBLName2 on $now_date"; if ($Use_Title == 1) { echo("$title\n"); } $sep = "\t"; //tabbed character for ($i = 0; $i < mysql_num_fields($result); $i++) { echo mysql_field_name($result,$i) . "\t"; } print("\n"); while($row = mysql_fetch_row($result)) { $schema_insert = ""; for($j=0; $j<mysql_num_fields($result);$j++) { if(!isset($row[$j])) $schema_insert .= "NULL".$sep; elseif ($row[$j] != "") $schema_insert .= "$row[$j]".$sep; else $schema_insert .= "".$sep; } $schema_insert = str_replace($sep."$", "", $schema_insert); $schema_insert = preg_replace("/\r\n|\n\r|\n|\r/", " ", $schema_insert); $schema_insert .= "\t"; print(trim($schema_insert)); print "\n"; } echo "\n...\n"; } if($DB_TBLName3) { $sql = "Select * from ".$DB_TBLName3." where execution_date >= '".$search_from_date."' and execution_date <= '".$search_to_date."' order by execution_date desc"; $result = @mysql_query($sql,$Connect) or die("Couldn't execute query:<br>" . mysql_error(). "<br>" . mysql_errno()); $Use_Title = 1; $title = "Report for $DB_TBLName3 on $now_date"; if ($Use_Title == 1) { echo("$title\n"); } $sep = "\t"; //tabbed character for ($i = 0; $i < mysql_num_fields($result); $i++) { echo mysql_field_name($result,$i) . "\t"; } print("\n"); while($row = mysql_fetch_row($result)) { $schema_insert = ""; for($j=0; $j<mysql_num_fields($result);$j++) { if(!isset($row[$j])) $schema_insert .= "NULL".$sep; elseif ($row[$j] != "") $schema_insert .= "$row[$j]".$sep; else $schema_insert .= "".$sep; } $schema_insert = str_replace($sep."$", "", $schema_insert); $schema_insert = preg_replace("/\r\n|\n\r|\n|\r/", " ", $schema_insert); $schema_insert .= "\t"; print(trim($schema_insert)); print "\n"; } } ?> have anyone any idea? Similar TutorialsI want to have a search product feature, but I would like members to be able to search multiple fields in one go i.e. product_code, Product_name in one MySQL query. The thing is, members have to be logged on, so the query must also only show results relating to that specific member, via the session[member_ID], my current query for listing products for that specific member is : Code: [Select] $sql = "SELECT productId, productCode, image, name, price, stock_level FROM product_inventory WHERE memberr_ID = '" . $_SESSION['SESS_mem_ID'] . "; How would I change the above into a search query to search for productcode, productname and still only show results beloging to this member using the session data ? all help appreciated i have a mysql table which contains name like mid mname 101 AAA 102 BBB 103 CCC now i have to print this name in a html table like AAA, BBB, CCC i am getting this by while loop in a variable but when loop changes then value also change so please tell me how i get this only in one variable & print Hey in my edit page i have 2 radio buttons in my form and i need to make sure the same value is still selected how can i do that? thanks I have problem that only last text field is updated, how should I fix this? Here's the code <?php if(isset($_POST["update"])){ mysql_query("UPDATE categories SET name_category = '".$_POST['category']."' WHERE ID= ".$_POST['currentCat']." ") or die(mysql_error()); mysql_query("UPDATE podkategorije SET name_subcategory = '".$_POST['subcategory']."' WHERE id_subCat= ".$_POST['currentSubCat']." ") or die(mysql_error()); } ?> <form action="" method="post" > <?php //creating texfields from db $query = "SELECT k.ID, k.name_category, pk.name_subcategory, pk.id_subCat FROM `categories` AS k JOIN `subcategories` AS pk ON pk.id_mainCat = k.ID"; $result = mysql_query($query) or die(mysql_error()); $currentCat = false; while($row = mysql_fetch_array($result)) { //so it doesn't repeat itself if($currentCat != $row['ID']) { //display of main Categories ?> <ul> <li> <br/><input name="categories" type="text" value="<?php echo $row['name_category']; ?>" /> </li> </ul> <? $currentCat = $row['ID']; } //display subcategories ?> <input name="subcategories" type="text" value="<?php echo $row['name_category']; ?>" /><br/> <input type="hidden" name="currentCat" value="<?php echo $row['ID']; ?>" /> <input type="hidden" name="currentSubCat" value="<?php echo $row['id_subCat']; ?>" /> <? } ?> <br /> <input type="button" value="Back" onClick="history.go(-1);return true;"> <input type="submit" value="Update" name="update"/> </form> Hi all, I am trying to make a members details section that can be updated. I want to be able to "SELECT * FROM users WHERE email='$email'" and then show the values that can be changed in a html drop down box with the selection that was made when the user registered already selected="selected"; You will be able to see what I am attempting to do below. <?php $sql = "SELECT manufacturer FROM table1 WHERE email=$'email'"; $result = mysqli_query($cxn,$sql); $row = mysqli_fetch_assoc($result); foreach manufacturer in table1 { if table1.manufacturer = table2.manufacturer { echo '<option name="manufacturer" selected="selected" value"$row['manufacturer']"</option>'; } else { echo '<option name="manufacturer" value"$row['manufacturer']"</option>'; } } ?> I have the following html code and php code. I would like clients to select which fields in the mysql database they would like to have appear in the resultant table by checking the text box next to the field in the html form. For instance they should be able to select first name and city and have only those columns appear in the resultant table. # <html> # <head> # <title>Search Clients Database</title> # </head> # # <body> # <h1>Clients Database Search Page</h1> # <form action="searchclients.php" method="post"> # <table width="100%" border="1" cellspacing="1" cellpadding="1"> # <tr> # <th width="37" scope="col"> </th> # <th width="114" scope="col">Fields</th> # <th width="169" scope="col">Filter Value</th> # <th width="1157" scope="col"> </th> # </tr> # <tr> # <td><input type="checkbox" name="idc" id="idc"></td> # <td>ID</td> # <td><input type="text" name="ID" id="ID"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="fnc" id="fnc"> # </label></td> # <td>First Name</td> # <td><input type="text" name="FirstName" id="FirstName"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="lnc" id="lnc"> # </label></td> # <td>Last Name</td> # <td><input type="text" name="LastName" id="LastName"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="cc" id="cc"> # </label></td> # <td>City</td> # <td><input type="text" name="City" id="City"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="pc" id="pc"> # </label></td> # <td>Province</td> # <td><select name="Province" id="Province"> # <option selected> </option> # <option>KZN</option> # <option>North West Province</option> # <option>Gauteng</option> # <option>Free State</option> # <option>Mpumalanga</option> # <option>Eastern Cape</option> # <option>Limpopo Province</option> # <option>Northern Cape</option> # <option>Western Cape</option> # </select></td> # <td> </td> # </tr> # </table> # <p> # <input type="submit" name="submit" value="Search" /> # </p> # </form> # # </body> # </html> # <?php # mysql_connect ("localhost", "username","password") or die (mysql_error()); # mysql_select_db ("clients"); # # $fn = $_POST['FirstName']; # $ln = $_POST['LastName']; # $city = $_POST['City']; # $prov = $_POST['Province']; # # $idc = $_POST['idc']; # $fnc = $_POST['fnc']; # $lnc = $_POST['lnc']; # $cc = $_POST['cc']; # $pc = $_POST['pc']; # # ?> # # <html> # <body> # # <table border=1> # <tr> # <th>ID</th> # <th>First Name</th> # <th>Last Name</th> # <th>City</th> # <th>Province</th> # </tr> # # <?php # $sql = mysql_query("select ID, FirstName, LastName, City, Province from clients where FirstName like '%$fn%' and LastName like '%$ln%' and City like '%$city%' and Province like '%$prov%'"); # # while ($row = mysql_fetch_array($sql)){ # # $id = $row['ID']; # $fname = $row['FirstName']; # $lname = $row['LastName']; # $city = $row['City']; # $prov = $row['Province']; # # ?> # # <tr> # <th><?php echo $id;?></th> # <th><?php echo $fname;?></th> # <th><?php echo $lname;?></th> # <th><?php echo $city;?></th> # <th><?php echo $prov;?></th> # </tr> # # <?php } //this ends the if?> # # </table> # </html> what's wrong with this ? i can't upload Apologies... I'm a noob trying to reverse engineer code. I have a script that can retrieve values from a MySQL database (based on a session login that recognizes which user you are) and places those values into the text fields of a form, pre-populating the form. Code: [Select] <?php mysql_connect('localhost', 'db', 'password'); mysql_select_db('users'); $id = mysql_real_escape_string($_SESSION['userid']); $select = "SELECT * FROM user_info WHERE md5(Member_No) = '$id';"; $query = mysql_query($select); if ($row = mysql_fetch_array($query)) { ?> <form> FIRST name: <input type="text" name="First" value="<?php echo stripslashes($row['First']); ?>" LAST name: <input type="text" name="Last" value="<?php echo stripslashes($row['Last']); ?>" > Birthday: <input type="text" name="Birthday" value="<?php echo stripslashes($row['Birthday']); ?>" <input type="submit" value="Submit"> </form> <?php } ?> I want to be able to connect to that same database and update ALL the values (3 in the example above) with the single click of the form button. I think I was told some time ago that it's not possible to update all the values at once? Thanks, ~Wayne Hey Everyone, I'm creating a site that will show images uploaded for certain days working on a job site. Kind of a day-to-day photo journal for the customer. On the site, the user gets here, sees 3 large images, and a series of thumbnails if more than 3 images exist for that day (works fine). However, underneath that I want to display a 3-4 column setup of "Archived Dates" that provide a link to the images of the other dates. I have this working correctly, but the results are displayed as follows: Date 1: Date 2: Date 3: etc.... I want them to display like this; Day 1 Day 4 Day 2 Day 5 Day 3 Day 6 and so on..... in a 3 column format. Here is the code I have right now just looping through to display these link results, not the rest of the page. I am trying to do it tableless right now, but if that isn't the right way to go, please let me know. Thanks to anyone in advance, Nick $SQLRowe = "SELECT DISTINCT RoweImgDate from tblRowe WHERE RoweImgDate !='" . $_GET['date'] . "' Order by RoweImgDate DESC Limit 0, 30"; //echo $SQLRowe; $rsSQLRowe = mysql_query($SQLRowe); <span class="rowe">Archived Photos:</span><br/> <div id="archive"> <?php while($row = mysql_fetch_array($rsSQLRowe)){ //echo "<a href='index.php?id="' . $row[RoweImgID] . '"' class='link'>$row[RoweImgDate]</a></br>"; echo "<div id='archivedates'>"; echo "<a href='index.php?date=" . $row[RoweImgDate] . "' class='link'>$row[RoweImgDate]</a>"; echo "</div>"; //echo "<img src='images/$row[RoweImage]'/><br/>"; //echo "<span class='FeatDesc'><p>$row[RoweImgDesc]</p></span><br/>"; } ?> </div> Hi, here's my problem: I am trying to make a simple online buying website and I want to display a table with all the fields for each item. So I got that part down which is to just use mysql_fetch_assoc("SELECT * FROM myTable") and use the html table tags stuff, but now I want to display my images in the table, so here's my code to display my mysql database table in html's table tag along w/ php: <html> <head> <title>My Online buying website project</title> </head> <body> <?php mysql_connect("localhost","root"); mysql_select_db("myTable"); $imagesArray=array("Apple_iPhone3GS.jpg","Apple_iPhone4.jpg","product3.jpg","product4.jpg","product5.jpg"); $result=mysql_query("SELECT Name, Manufacturer, Price, Description, SimSupport FROM myTable"); if(mysql_num_rows($result))//if there is at least one entry in bellProducts, make a table { print "<table border='border'>"; print "<tr> <th>Name</th> <th>Manufacturer</th> <th>Price</th> <th>Description</th> <th>SimSupport</th> </tr>"; //NB: now output each row of records while($row=mysql_fetch_assoc($result)) { extract($row); print "<tr> <td>$Name</td> <td>$Manufacturer</td> <td>$Price</td> <td>$Description</td><td>$SimSupport</td> </tr>"; }//END WHILE }//END IF ?> </table> </body> </html> *So how do I go about adding my images in this table? Hi, i don't understand how to show ENUM type data in php. Now i use: <input type="text" name="type" style="width: 362px; height: 35px;" value="<?php echo $type; ?>"> But it print only the value that is set to topic i see. How to display all ENUM values in radio buttons and that checked button would be that one from the topic. For example: ENUM ('city','town','vilage') . I select town when writing a topic. And when i edit data i see 3 radio buttons and TOWN would be checked. Thank you for answer. p.s maybe someone know a good site where is a tutorial how to show enum in php Hi, I am having difficulty deleting rows in my table using check boxes. I have all the check boxes displaying for each row and a delete button below the table. However when i click on the delete button it isnt deleting the checked rows, i have displayed the code below that i am using; Code: [Select] <?php $result = mysql_query("SELECT * FROM contact ORDER BY msg_id ASC"); echo "<table border='1'> <tr> <th>Delete</th> <th>Message ID</th> <th>Name</th> <th>Email</th> <th>Message</th> </tr>"; while($row = mysql_fetch_array($result)) { ?> <td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<?php echo $row['del_id']; ?>"></td> <?php echo "<td>" . $row['msg_id'] . "</td>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['email'] . "</td>"; echo "<td>" . $row['msg'] . "</td>"; echo "</tr>"; } echo "</table>"; ?> <td colspan="5" align="center" bgcolor="#FFFFFF"><input name="delete" type="submit" id="delete" value="Delete"></td> </tr> <?php // Check if delete button active, start this if(isset($_GET['delete'])) { for($i=0;$i<$count;$i++){ $del_id = $checkbox[$i]; $sql = "DELETE FROM contact WHERE id=".$_GET['del_id']; $result = mysql_query($sql); } // if successful redirect to delete_multiple.php if($result){ } } mysql_close(); ?> Please help Thank you I am working on this page concept of pedigree. How can I foreach or create a loop that can get 6 levels of parents, for example name= 1 ; nfather=2; nmohter = 3 and the loop continue to get or consider nfather as a name = 2 and get his parents. this is my table any help plees
CREATE TABLE horsetest01 (
CREATE INDEX horsetest01_index02
Hi, I have a 'pre-release' sign up form, where people put in their email address and name, and it transmits to a MYSQL database. That's working fine. What I want to do is to have a piece of text that reads: (insert pre release amount of signer upperers here) have signed up. Will you? I'm guessing that the amount of rows on the table, would end up being the number. How do I do this? Thanks, Jack I have the following code that searches my database and displays results in a table: $fields = array("field1", "field2", "field3") $cols = implode (', ', $fields); $result= mysql_query (" SELECT $cols FROM tablename WHERE ................... "); if (!$result) {die('Could not search database: ' . mysql_error());} if($result) { if(mysql_num_rows($result) == 0) { return "Sorry. No records found in the database"; } else { $table = "<table border='1' cellpadding='5' cellspacing='5'>"; while($arr = mysql_fetch_array($result, MYSQL_ASSOC)) { $table .= "\t\t<tr>\n"; foreach ($arr as $val_col) { $table .= "\t\t\t".'<td>'.$val_col.'</td>'."\n"; } $table .= "\t\t</tr>\n"; } $table .= "</table>"; echo $table; } mysql_free_result($result); } As you can see each of the MySQL table fields specified by $fields is displayed in a new column in the html table. I want to change this so that e.g. "field3" is displayed in a new row instead. So, instead of the html table looking like: | "field1-result1" | "field2-result1" | "field3-result1" | | "field1-result2" | "field2-result2" | "field3-result2" | | "field1-result3" | "field2-result3" | "field3-result3" | I want it to look like: | "field1-result1" | "field2-result1" | | "field3-result1" | | | "field1-result2" | "field2-result2" | | "field3-result2" | | | "field1-result3" | "field2-result3" | | "field3-result3" | | I guess this is quite straightforward, but I can't work it out! Pls help! Thanks. Here is one of those purely conceptual questions, which involves no code. I'm trying to create a select query which among other things, allows a user of a website to search other members who fall within a particular age range. I have a column in the table where the members' information are stored which records their ages. Which brings us to the problem. On any given day, a member's age may increase by one year compared to what it was the previous day, hence the need to update this column periodically. I can't think of any way to automatically update this column on a daily basis. So my solution is to run an initial update query every time a member tries to search other members based on age, which updates the age column for all other members, before running the select query which eventually retrieves the desired age range. This leads to the second problem. Imagine there are thousands of users using the website. At any given instance, there could be hundreds of members, trying to search others based on age. This means hundreds of users will be updating a single column (the age column) in one table at the same time. Is this feasible? Can it cause the server to crash? Or is there really a more reasonable way to do all of this? Thank you all for taking your time to read this. Appreciate any responses. First, Happy X-Mas to all! I want to do a form for entering dog-show results. On first site of form the user is able to enter how many dogs for each class (there are youth and open class for example) he want to insert. So on second site there will be formfields for every dog in every class, created with php. User should be able to take dogname from jquery autocomplete - this works for multiple fields. But i need the according id to the dogname - and actual only one id is given - the first one is changing by choosing from autocomplete in next fields... Data came from a json array like
0: {dogidindatabase: "9892", value: "Excalibur Khali des Gardiens de la Cour ",…}
dogidindatabase: "9892"
label: "<img src='../main/img/female.png' height='20'>Excalibur Khali des Gardiens de la Cour "
value: "Excalibur Khali des Gardiens de la Cour "
1: {dogidindatabase: "15942", value: "Excalibur from Bandit's World Kalli",…}
dogidindatabase: "15942"
label: "<img src='../main/img/male.png' height='20'>Excalibur from Bandit's World Kalli"
value: "Excalibur from Bandit's World Kalli"
all i need is inside ...the script in form is
<script type='text/javascript'>
//<![CDATA[
$(function() {
$(".dog").autocomplete({
source: "xxx.php",
minLength: 3,
select: function(event, ui) {
$('#dogidindatabase').val(ui.item.dogidindatabase);
}
});
$["ui"]["autocomplete"].prototype["_renderItem"] = function( ul, item) {
return $( "<li></li>" )
.data( "item.autocomplete", item )
.append( $( "<a></a>" ).html( item.label ) )
.appendTo( ul );
};
});
//]]>
</script>
and the form looked like
<input type='text' name='dog' class='dog' value='".strip_tags(${'dogname' .($countbabymale+1)})."' size='35' data-required='true' /><br>
<input class='readonly' readonly='readonly' type='text' id='dogidindatabase' name='dogidindatabase' size='5' />
I changed in script and form id='dogidindatabase' to class, but it doesn´t work.
Testsite is under http://www.wolfdog-d...how.php?lang=de (only german at first) - Insert a number (higher than 1) under baby 3-6 (first box, the others are not working for testing); submit to get to page 2; you see all post-variables (for testing).
Try to insert a dogname in first input-field (choose "exc") and insert first one - the id is under dogname. In second dogname inputfield you can choose second dog - the id for the first one changes to id of seond one ....
How can i get both for every dog?
i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
Hi, I want to download more than one file at a time. But now I can't download any of them that way. I want to solve it the way it is in gmail, i.e. I download either 1 file separately or if there are two or more, it is zipped. Thanks in advanced, T
<!-- images -->
<section class="card shadow mb-3">
<form action="index.php?lg=<?php echo $lng; ?>&c=text" method="post" enctype="multipart/form-data">
<header class="card-header navbar-custom"><h3 class="text-center"><?php echo $new_document_h3_images; ?></h3></header>
<div class="row">
<?php
$files = glob('./tekstovi/' . $user_txt_year . '/' . $user_txt_nr . '/' . $text_page . '/' . $user_txt_year . '_' . $user_txt_nr . '_' . $text_nr . '_' . $text_page . '_img_' . '*.{doc,docx,odt,pdf,jpg,JPG,jpeg,png,gif,psd,eps,ai,tiff,tif}', GLOB_BRACE);
$dir = './tekstovi/' . $user_txt_year . '/' . $user_txt_nr . '/' . $text_page . '/';
//$index = '0';
for ($i = 0; $i < count($files); $i++) {
$image = $files[$i];
?>
<div class="col-sm-4 py-2">
<div class="card h-100">
<input type='checkbox' style="position: absolute; left: 10px; top: 10px;" name='boxes[]' value='<?php echo basename($image); ?>' />
<?php echo '<img class="img-fluid" src="' . $image . '" alt="" />'; ?>
<div class="card-body">
<p class="card-title"><?php echo basename($image); ?></p>
</div>
<div class="card-footer"><?php
$stmt_captionreload = $pdo->prepare('SELECT * FROM `sn_images`
WHERE sn_images_text_id = :sn_images_text_id
AND CONCAT(`sn_images_filename`,`sn_images_basename`) = "' . basename($image) . '"
');
$stmt_captionreload->bindValue(':sn_images_text_id', $text_id);
$stmt_captionreload->execute();
$row_captionreload = $stmt_captionreload->fetch(PDO::FETCH_ASSOC);
$one_caption = stripslashes($row_captionreload['sn_images_caption']);
?>
<small class="text-muted"><?php echo $one_caption; ?></small></div>
</div>
</div>
<?php }?>
</div><!-- row-->
<button type="submit" name="download" value="1" class="btn btn-success float-right">Download selected</button>
<input type = 'hidden' name="sn_text_nr" value="<?php echo $text_nr; ?>">
</form>
<?php
if ($_POST['download'] == '1') {
$index = '0';
echo 'Total count: ' . count($_POST['boxes']) . '';
while ($index < count($_POST['boxes'])) {
if (isset($_POST['boxes'][$index])) {
$file = $_POST['boxes'][$index];
$path = $dir . $file;
header('Content-Description: File Transfer');
header('Content-Type: application/octet-stream');
header('Content-Disposition: attachment; filename=' . basename($path));
header('Content-Transfer-Encoding: binary');
header('Expires: 600');
header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
header('Pragma: public');
header('Content-Length: ' . filesize($path));
flush();
readfile($path);
} else {
}
$index++;
}
}
?>
</section>
<!-- /. images -->
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