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PHP - Ajax Menu To Retrieve Mysql Records
I wonder whether someone can help me please.
I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page. Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be: Code: [Select] <?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?> The problem I have, is that when I try and run this, I receive the following error: Quote Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";. I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem. I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong. Many thanks and kind regards Similar TutorialsI have a db with 4 tables: table_A table_B table_C table_D table_A contains personal info about users and has a primary id field called uid. Each of tables B, C and D is linked to table_A via the foreign key f_uid. table_A also has a column called signed_up which can contain a 1 or a 0. I want to pull up lots of info from each of the 4 tables for every user, as long as the signed_up column in table_A contains a 1. So, I've made a query like this: Code: [Select] $q = "SELECT table_A . col_1 , table_A . col_2 , table_A . col_3 , table_B . col_1 , table_B . col_2 , table_B . col_3 , table_C . col_1 , table_C . col_2 , table_C . col_3 , table_D . col_1 , table_D . col_2 , table_D . col_3 FROM table_A, table_B, table_C, table_D WHERE table_A . uid = table_B . f_uid AND table_A . uid = table_C . f_uid AND table_A . uid = table_D . f_uid AND table_A . signed_up = '1' ORDER BY table_A . some_col "; $r = mysqli_query($dbc, $q); The query is executing but it's also pulling up records where signed_up in table_A contains a '0'. Can anyone see a logical flaw in my query?... Alternatively, is there an easier way I could build the query? TIA This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=311502.0 Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me Hello everybody!
I am trying to figure out how to obtain all the data related to a key, but I've got no results so far and I am becoming really frustated, let's see if any of you could help me out with this.
Imagine I have a table with several columns, but we bother about two of them, let's say we have serial numbers of some product, on the left Incoming serial number (we can repair or swap the unit), on the right the outcoming serial number (same if we have repared the unit, different if we swap it for another unit).
Then we have, for example:
A -> A (Unit A enters and we repaired it)
A -> B (Same unit came another day for some reason and we couldn't repair it, so we swap it by giving B to the customer)
B -> C (Unluckily B was defective so we have to change it again)
C -> C (C had another problem and we repaired it)
We have that in the database from different days and the such, so now, we want to know the historical and we know "C". If we perform a SELECT * FROM... WHERE incoming/outcoming serial number = "C" we'll get:
B -> C
C -> C
So we should seek now for B and keep going... but I cannot proceed correctly, 'cause if I SELECT using B I'll get again B -> C (and A -> B, what I want), but when do I know I have to finish? How could I implement this as a function or whatever? showing every not repeated line from the beginning.
Could your minds help mine? Thank you very much in advance!.
I want to use session to do a query and will I be able to do this? I have a session that was gathered from login and now i was to use this session to do a query If Yes, How? Hi, Im trying to retrieve HTML from a mysql database but nothing i've tried seems to work. The HTML im trying to retrieve is an iframe with a link and styles (code from amazon associates). Im trying to display links to specific products on amazon from the product page on my site. All data about the product is retrieved from the database so i have code to select the amazon link row in my database table but i cant get it to display. It says the html isnt a string so i cant echo it, fair enough. I have tried using the following code: Code: [Select] $get_buylink_sql = "SELECT mobo_buylink FROM mobo WHERE mobo_id = $mobo_id"; $get_buylink_res = mysqli_query($mysqli, $get_buylink_sql) or die(mysqli_error($mysqli)); while ($buylink = mysqli_fetch_array($get_buylink_res)) { echo"<iframe src=\"".$buylink."\" style=\"width:120px;height:240px;\" scrolling=\"no\" marginwidth=\"0\" marginheight=\"0\" frameborder=\"0\"></iframe>"; } mysqli_free_result($get_buylink_res); mysqli_close($mysqli);I have also tried putting the whole iframe code in the database which didnt work either. The mobo_id variable works fine for retrieving the rest of the data and i need to get the amazon link from the same record. I hope i've put this in a way you can understand, but if not i'll try and explain better and give you a link to my site if needed. Thanks, Alex Hi folks, Complete No0b here when it comes to PHP and MySQL. I am in the middle of creating a PHP website. What I want to do is have the contents of a page in a MySQL table and have PHP gather the page content and display it on the page. Here is what I have in my home.tpl file: Code: [Select] <?php $query = "SELECT home, FROM $database_name"; $result = mysql_query($query); ?> And, in my index.php I simply call that home.tpl to display the data by using: Code: [Select] <?php include 'templates/default/home.tpl'; ?> Now, all I get is a blank page on index.php. Is there something else I should be doing? Remember, I am a complete No0b! Thanks folks. I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks Hello I am having a trouble with creating page that will search records from database using dropdown menu, when I select a category, for example By: Gender and after clicking the Submit button, there is no records displays but when I select By: ALL, the records displays. What's wrong with my code? I'm sorry I'm fairly new to PHP. Here is my code: Code: [Select] <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <title>Search</title> </head> <body> <form action="" method="POST" name="records"> <label>SEARCH: </label> <select name="select"> <option value="all" selected="selected">ALL</option> <option value="year">By Year</option> <option value="gender">By Gender</option> <option value="course">By Course</option> </select> <input name="submit" value="GO" type="submit" /> <br /> </form> </body> </html> <?php include('collegeinfo_connect.php'); mysql_connect("$server", "$user", "$pass")or die("cannot connect"); mysql_select_db("$db")or die("cannot select DB"); if (isset($_POST['submit'])) { $selection = $_POST['select']; if($selection == "all") $query = "SELECT * FROM collegeinfo_tbl"; else if($selection == "year") $query = "SELECT * FROM collegeinfo_tbl WHERE Year='year'"; else if($selection == "gender") $query = "SELECT * FROM collegeinfo_tbl WHERE Gender='gender'"; else if($selection == "course") $query = "SELECT * FROM collegeinfo_tbl WHERE Course='course'"; } if (isset($query)) { $result = mysql_query($query) or die(mysql_error()); echo "<br /> <br />"; echo "<table border='1' cellpadding='10'>"; echo "<tr> <th>ID Number:</th> <th>First Name:</th> <th>Last Name:</th> <th>Gender:</th> <th>Year:</th> <th>Course:</th> </tr>"; while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo '<td>' . $row['ID'] . '</td>'; echo '<td>' . $row['FirstName'] . '</td>'; echo '<td>' . $row['LastName'] . '</td>'; echo '<td>' . $row['Gender'] . '</td>'; echo '<td>' . $row['Year'] . '</td>'; echo '<td>' . $row['Course'] . '</td>'; echo "</tr>"; } echo "</table>"; } ?> Hello all, I am new to php coding and have a couple of problems with editing records in my database! I have two files below one test.php and edit.php. In the test.php the code outputs the records into a table. The problem is with the edit link as when it is selected I wish to be able to edit a record by a form, which is on edit.php. I am trying bring up the movie's information on the form to be edited. Currently on the form i get; Quote Movie: movie Gen Genre Year: year Any ideas how I can edit the record and then return to the test.php page? Code: [Select] test.php <html> <body style="background-color:#669999;"> <table width="490"border=0><tr> <td colspan="2" style="background-color:#FFA500;"> <div id="header" <h3 style="color:black">This is my first web-page! Below is a database of some of my favourite movies! </h3> </td> </tr> <?php //connecting to server $con = mysql_connect("localhost","root","NYOXAkly"); if (!$con) { die('could not connect: ' . mysql_error()); } //selecting movie database mysql_select_db("my_mov",$con); //Check if add button is active, start this if(isset($_REQUEST['add'])) { echo "<meta http-equiv=\"refresh\"content=\"0;URL=form.php\">"; } $result = mysql_query("SELECT * FROM Films ORDER BY filmID"); ?> <!-------------------------------creating table------------------------------------------------------------------------------------> <table width="490" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <form name="test1" method="post" action="test.php"> <table width="490" border="10" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"><tr> <td bgcolor="#FFFFFF"></td> <td align="center" colspan="6" bgcolor="#FFFFFF">Movie Database</td></tr> <td align="center" bgcolor="#FFFFFF">filmID</td> <td align="center" bgcolor="#FFFFFF">Movie</td> <td align="center" bgcolor="#FFFFFF">Genre</td> <td align="center" bgcolor="#FFFFFF">Year</td> <td align="center" bgcolor="#FFFFFF">Edit</td> <td align="center" bgcolor="#FFFFFF">Delete</td> </tr> <?php while($rows=mysql_fetch_array($result)) { ?> <tr> </td> <td bgcolor="#FFFFFF"><? echo $rows['filmID']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['movie']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['genre']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['year']; ?></td> <td bgcolor="#FFFFFF"> <a href="edit.php?filmID=<?php echo 'filmID';?>">Edit</a> <td bgcolor="#FFFFFF"> <a href="delete.php?filmID=<?php echo 'filmID';?>">Delete</a> </td> </tr> <?php } echo print_r(error_get_last()); mysql_close(); ?> <!--add button--> <tr> <td colspan="15" align="left" bgcolor="#FFFFFF"> <input name='add' type="submit" filmID="add" value="Add A New Record" action="form.php?"> </td> </tr> </table> </form> <br/> By C.M.D.W <br/> <?php echo date("Y/m/d") . "<br />"; ?> </body> </html> Code: [Select] edit.php <html> <body style="background-color:#669999;"> <!------------------Creates a form ----------------------> <br /> <form action="" method="post"> <fieldset> <legend>Enter your movies into database here!</legend> Movie: <input type ="text" name="movie" value="<?php echo 'movie';?>"> <br /> Gen <input type ="text" name="genre" value="<?php echo 'genre';?>"/> <br /> Year: <input type ="text" name="year" value="<?php echo 'year';?>"/> <br /> <input type="submit" name="name" value="Submit" /> </fieldset> </form> <?php //connecting to server $con = mysql_connect("localhost","root","NYOXAkly"); if (!$con) { die('could not connect: ' . mysql_error()); } //selecting movie database mysql_select_db("my_mov",$con); if (isset($_POST['submit'])) { // confirm that the 'id' value is a valid integer if (is_numeric($_POST['filmID'])) // get form data $filmID = $_POST['filmID']; $movie = mysql_real_escape_string(htmlspecialchars($_POST['movie'])); $genre = mysql_real_escape_string(htmlspecialchars($_POST['genre'])); $year = mysql_real_escape_string(htmlspecialchars($_POST['year'])); // check that fields are filled in if ($movie == '' || $genre == '' || $year == '') { // generate error message $error = 'ERROR: Please fill in all required fields!'; } else { // save the data to the database } mysql_query("UPDATE players SET movie='$movie', genre='$genre', year='$year' WHERE filmID='$filmID'") or die(mysql_error()); // once saved, redirect back to the view page header("Location: test.php"); } } if (isset($_GET['filmID']) && is_numeric($_GET['filmID']) && $_GET['filmID'] > 0) { // query db $id = $_GET['filmID']; $result = mysql_query("SELECT * FROM Films WHERE filmID=$FilmID") or die(mysql_error()); $row = mysql_fetch_array($result); // check that the 'id' matches up with a row in the databse if($row) { // get data from db $movie = $row['movie']; $genre = $row['genre']; $year = $row['year']; }} ?> <br/> <br/> <a href="test.php">Return To Home Page</a> <br/> <br/> By C.M.D.W <br/> <?php echo date("Y/m/d") . "<br />"; ?> </body> </html> Thanks Chris Hi guys, I am trying to UPDATE some records on a mySQL database but can't seem to find out why it is not working. This is my code. Code: [Select] <?php $latitude = $_POST['lat_location']; $longitude = $_POST['long_location']; $unique_ID = $_POST['unique_ID']; include('connect2.php'); $query = mysql_query("SELECT * FROM user_location WHERE unit = '$unique_ID'"); $numrows = mysql_num_rows($query); if ($numrows == 1) { $query2="UPDATE user_location SET lat = '$latitude', long = '$longitude' WHERE unit = '$unique_ID'"; mysql_query($query2); $test = "matches"; } else { mysql_query("INSERT INTO user_location VALUES ('','$unique_ID','$latitude','$longitude')"); $test = "not match"; } echo $test . "<br />"; echo $numrows; ?> The script receives the data via the POST method and assigns it to variables. Then I query the database for one of those variables and check how many results are found. If 0 results are found then a new record is created on the database, but if there is 1 record found then the record that is found has to be UPDATED. When the result is 0 the scripts creates the new record fine, but if the result is 1 it doesn't update. I just can't figure out why. Any help will be greatly appreciated. Thanks in advanced, I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> I m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php include("db.php"); $query=mysql_query("select * from table where id='3' "); $row=mysql_fetch_array($query); $r=$row['image']; header("content-type:image"); echo $r; ?>" i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance... Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago. I started out with some tutorials and feeling im getting the hang of it. Enough of me, lets get to the point: <?php $con = mysql_connect('localhost',$user,$pass)or die(mysql_error()); $selectdb = mysql_select_db($selectdb)or die(mysql_error()); $sql = "SELECT * From table"; $result = mysql_query($sql); $num = mysql_num_rows($result); $myarray = array($result); $i =0; while ($i < $num){ echo $myarray[$i]; $i++; } ?> Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results. The output in the browser is this: Resource id #3 I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple". Thanks in advance! Dan-Levi 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, My output in html is : Code: [Select] <uL><li id="B1"></li> <li id="B2"></li> <li id="B3"></li> <li id="B4"></li> <li id="B5"></li> <li id="B6"></li> <li id="B7"></li> <li id="B8"></li> <li id="B9"></li> <li id="B10"></li> <li id="B11"></li> <li id="B12" class="active"></li> <li id="B13" class="no"></li> <li id="B14" class="no"></li> <li id="B15" class="no"></li> <li id="B16" class="no"></li> <li id="B17" class="no"></li> <li id="B18" class="no"></li> <li id="B19" class="no"></li> <li id="B20" class="no"></li> </ul> If MySQL query result is equal to `6`, then `<li>` tag with `id` equal to "`B12`" should have class "`active`". All the `<li>` elements occuring after this active element should have class "`no`". This shows images of horizontal rating between `0` and `10` and between 0.5 Example : 0 .5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 In the example above elements from `0` to `5` will be blue, element `6` will be white and elements from `7` to `10` are black. How could I generate this using PHP and/or MySQL? Thanks, XXXX chirstmas XXXX hi i have a mysql table with upddate and updtime field as follows: upddate updtime 2011-02-25 11:03:05 2011-01-28 08:01:09 2011-02-20 07:00:08 2011-02-15 06:10:02 2011-02-05 10:21:04 2011-02-09 10:20:25 the data types for upddate is mysql date and for updtime its mysql time In my php page i use Code: [Select] <?php echo $date = date("d M Y",strtotime($row_rs1['upddate'])); ?> to retrieve date and the results displays as 25 Feb 2011. And for time right now im using Code: [Select] <?php echo $row_rs1['updtime']; ?> which displays time as 11:03:05 I want to display time as HH:MM only, how can i do it? Hello, This is driving me crazy, as I cannot get it working. I have a database that has a table called 'categories'. I need to retrieve all categories with category_id != '1,2,3'. For this, I can this query: Code: [Select] $q = "SELECT * FROM categories WHERE category_id NOT IN (1,2,3)"; The query executes fine. I need this categories retrieved, to be passed to a Smarty Template Engine powered page. I have: Code: [Select] $q = "SELECT * FROM categories WHERE category_id NOT IN (1,2,3)"; $res = mysql_query($q); while($row=mysql_fetch_array($res)) { Assign('categories',$row); } But when retrieving the array in the Smarty page, with a loop, I only see the last item of the array (the last category). If I go back to the while loop, and add a Code: [Select] print_r($row);, I see two arrays (I have two categories). But the question is: How can I work with those two categories, as I need to use them individually outside the while loop? Any help is appreciated. Best Regards, Richi I'm sorry to be back so soon, but I'm up against another mystery. I'm using the code below to enter a bunch of css data from a spreadsheet into a mysql table. I think the data file is OK. The array created by the script checks out with print_r. (There are many more records than shown. I truncated it to save space.) The problem is that I get this error regarding my sql statement, not the data or anything else: Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'check, name, phone, email, entry_fee, print_fee, image_name, description, med...' at line 1 in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php:242 Stack trace: #0 /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php(242): PDO->prepare('INSERT INTO tbl...') #1 {main} thrown in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php on line 242 I've typed it in a dozen times to make sure there are no errors and keep getting the same error. I tried running a test file and gradually increasing the number of placeholders and at some point I always end up getting the same error, I can delete the most recent addition and it works again. Then I can add another placeholder exactly as before and it works the second time. It feels like a ghost in the machine. Any idea what I am doing wrong? An I typing something I don't see?
<?php
require '__classes/Db.php';
$csvData = '1,FALSE,Carol Lettko,,,TRUE,FALSE,Carol_Lettko-DSC_3022.jpg,Baby Herons/Brickyard,photo,,,
,,,925-285-0320,cjl164@aol.com,,,Carol_Lettko-DSC_0164.JPG,Heron/Brickyard,photo,,,
,,,,,,,Carol_Lettko-IMG_5723.jpg,Kayaker/Brickyard,photo,,,
,,,,,,,,,,,,
2,FALSE,Louise Williams,,,TRUE,FALSE,Louise_Williams-BirdsOfAFeatherAOPR.jpg,Alligator with Words,Book Excerpt,,,
,,,510-232-9547,lkw@louisekwilliams.com,,,Louise_Williams-Hope-TheFairyChickenAOPR.jpg,Hope The Fairy Chicken,,,,
,,,The d exatrfrfvct/.*tygrvurr,,,,,,,,,
,,,,,,,,,,,,
3,TRUE,Dorothy Leeland,,lelanddorothy@gmail.com,TRUE,FALSE,DJ_Lee-bridge at dusk 700px width.jpg,Bridge,photo,,,
,,,,,,,DJ_Lee-friends 700px width.jpg,Friends,photo,,,
,,,,,,,DJ_Lee-hybiscus 700 px wide.jpg,Hibiscus,photo,,,
,,,,,,,,,,,,
4,FALSE,Rita Gardner,,,TRUE,FALSE,Rita_Gardner-Explosion - Gardner photo.JPG,Explosion,photo,,,
,,,,tropicrita@msn.com,,,Rita_Gardner-Ferry Point tables and chair - Gardner.JPG,Ferry Point Tables,photo,, ,
,,,,,,,Rita_Gardner-Forks - Gardner photo.JPG,Forks,photo,,,
,,,,,,,,,,,,
';
$lines = explode(PHP_EOL, $csvData);
$array1 = array();
foreach ($lines as $line) {
$array1[] = str_getcsv($line);
}
$stmt = $pdo->prepare("INSERT INTO tbl_person_data (number, check, name, phone, email, entry_fee, print_fee, image_name, description, medium, select, orient, site) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)");
foreach ($array1 as $row)
{
$stmt->execute('$row');
}
echo '<pre>';
print_r($array1);
echo '</pre>';
?>
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