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PHP - Looping Through A Joined Table Then Inserting Into Two Tables
No idea what I am doing here. I have a joined table which is:
$sql = "SELECT itemnum, image1, title, close, quantity, bidquantity, ". I need to loop through the table on buser and insert select data into two other tables. One of the tables is the control table which will only have one row per buser. The other table I guess you would call the data table of the two and may have more than one row depending. The catch is these will be linked together by an invoice number. Table A, the control table has a field for invoice number, table B does not. Of course once the data is inserted into the new tables they will be joined and from there on it should be easy. I just don't know now how to get there. I've thought about using an array to loop through, but I don't know enough about them to make any sense of this. Not even sure the group by may help me in doing this. Anyone have any ideas on this? I sure do need it and thanks in advance! Similar TutorialsHi all I have 2 tables: 1. is a tables full of images that have an id, a src and a title (images_table). 2. table is a list of records with an id, a description and an images_id (records_table) What I'm looking to do is: Loop through the records in the records_table and find the description, the description is something like this 'Welcome to Kansas'. Then, I want to loop through the images_table and find the associated image of Kansas, based on a MySQL LIKE statement. This is because the images_table has a title of 'Kansas', for example. Once this has been done, I then need to insert the id of Kansas into the images_id in the records_table a possible MySQL UPDATE. Has anyone got an idea how I could do this? Thanks Hi,
I have the following query
SELECT user_details.User_club_ID, user_details.fname, user_details.lname, user_details.email, user_details.club_No club.CLUBCODE, club.club_id FROM user_details, club WHERE club_id = $cid AND user_details.club_No = club.CLUBCODE AND user_status = 'active'";which I converted to a prepared statement as SELECT user_details.User_club_ID, user_details.fname, user_details.lname, user_details.email, user_details.club_No club.CLUBCODE, club.club_id FROM user_details, club WHERE club_id = ? AND user_details.club_No = club.CLUBCODE AND user_status = ?";Please note that user_status is a field in the table user_details. The original query (non -PDO) works correctly. I want to know if this is correct and that the comparison in the WHERE clause i.e. user_details.club_No = club.CLUBCODE is security safe. If not then how should this be modified. Also if there is a better way to write this statement, kindly show that as well. Thanks Thanks all ! Edited by ajoo, 11 December 2014 - 02:35 AM. Back with a new problem. I have 8 tables interconnected. Table#1 - Users user_id | name Table#2 - user_categories id | user_id | category_id Table#3 - user_cities id | user_id | city_id Table#4 - user_dates id | user_id | dates_available Table#5 - categories category_id | category_name Table#6 - cities city_id | city_name Table#7 - provinces province_id | province_name Table#8 - categories country_id | country_name Each user will have multiple categories, cities and available dates listed in these tables. I simply want to retrieve and list each user and their data on a page. Here's my query. $url_city = 1;
$url_category = 2;
$url_date = '2021-07-19';
$find_records = $db->prepare("SELECT user_categories.*, categories.*, user_cities.*, cities.*, provinces.*, countries.*, user_dates.*, users.* FROM users
LEFT JOIN user_categories ON users.user_id = user_categories.user_id
LEFT JOIN user_cities ON users.user_id = user_cities.user_id
LEFT JOIN user_dates ON users.user_id = user_dates.user_id
LEFT JOIN categories ON user_categories.category_id = user_categories.category_id
LEFT JOIN cities ON user_cities.city_id = cities.city_id
LEFT JOIN provinces ON cities.province_id = provinces.province_id
LEFT JOIN countries ON provinces.country_id = countries.country_id
WHERE user_cities.city_id = :city_id AND user_categories.category_id = :category_id AND user_dates.date_available = :date_available GROUP BY users.user_id");
$find_records->bindParam(':city_id', $url_city);
$find_records->bindParam(':category_id', $url_category);
$find_records->bindParam(':date_available', $url_date);
$find_records->execute();
$result_records = $find_records->fetchAll(PDO::FETCH_ASSOC);
if(count($result_records) > 0) {
foreach($result_records as $row) {
$user_id = $row['user_id'];
$name = $row['name'];
$country_id = $row['country_id'];
$country_code = $row['country_code'];
$country_name = $row['country_name'];
$province_id = $row['province_id'];
$province_code = $row['province_code'];
$province_name = $row['province_name'];
$city_id = $row['city_id'];
$city_name = $row['city_name'];
$category_id = $row['category_id'];
$category_name = $row['category_name'];
}
}
There are no errors but the above query would only return a single row with only 1 "user" despite having multiple users in the "users" table. If I remove the GROUP BY, then it'll return multiple rows of the same user instead of all the relevant users. So what do you think I am doing wrong with my query? Edited July 20 by imgroootThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=317690.0 Hi, I have a form that once submitted some of its result are stored in arrays, example: (The form has multipul lines with the same input names) <select name="product[]"> once submitted goes into $_GET['product'] if I do: // Product ID's foreach($_GET['product'] as $name => $prodvalue) { print "$name : $prodvalue<br>"; } the following is returned: 0 : 9 1 : 10 2 : 11 3 : 12 Aswell as the Product ID's I have 2 other form input structured the same way, so my question is how do I loop through each of the $_GET's ($_GET['product'], $_GET['linequantity'] and $_GET['lineprice']) to add each of them to multipul SQL table rows? Also there will be other records that need to be entered, but, these will be constant, so for instance, if 3 rows are to be added then the other records will be the same for each of the 3 rows. Please help me, I'm goin' nuts! B. i have some code which loops through a table, and displays the results. i have then added extra code so that it loops through another table, and prints out the reults, that are related to the first loop. so the page should look like menu submenu 1 submenu 2 submenu 3 menu 2 submenu 4 etc The problem is that the code only prints out the first row from the first loop and nothing else. <?php $list = "SELECT * FROM section_main"; $result = mysql_query($list) or die ("Query failed"); $numofrows = mysql_num_rows($result); echo "<table border='1' id='section_list'>"; echo "<tr><th>section_id</th><th>section_title</th></tr>"; for($j = 1; $j < $numofrows; $j++) { echo '<tr>'; $row = mysql_fetch_array($result); echo "<td>". $row['section_id'] . "</td><td>". $row['section_title'] . "</td>"; $query = "SELECT section_sub.section_sub_title, section_sub.section_title WHERE section_sub.section_title = " .$row['section_title']."ORDER BY section_sub_title"; $result2 = mysql_query($query) or die ("query failed2");//This part does not work $numofrows2 = mysql_num_rows($result2); for($i = 0; $i<$numofrows2; $i++){ $row2 = mysql_fetch_array($result2); echo '<tr>'.$row2['section_sub_title'].''; } } echo "</tr>"; echo '</table>'; ?> Any help on whats wrong would be great Hey guys, I am kind of new to php but I am stuck on this question. I have a mySQL database and table set up called members. I have some data in the table and I want to be able to add certain things to a paticular row in the table based off the ID number of the row. One of the values of the table is an auto incremeted ID number. I want to add a text value called message to a specified ID number. How do I go about doing that. I have this code already but it doesn't seem to work. Code: [Select] $sql = "INSERT INTO members (message) VALUES ('$_POST[message]',(SELECT id FROM members WHERE id='$_POST[id]'))"; Any Ideas? Thanks I am working on a asset project to insert, delete and view assets (items) that i owe. I am having trouble in the insert part. i have four tables department (DeptID, DeptName) DeptID is the primary key in this table. assetcategory (AssetCatID, AssestCategory) AssetCatID is the primary key in this table. asset (AssetID, AssetDescription, EmpID, AssetCatID, DeptID, Model, Maker, SerialNUm, DateAguired) AssetID is the primary key here. Employee (EmpID, FirstName, LastName) EmpID is the key here. and this is the code i have to insert using php. Code: [Select] <?php if(isset($_POST['submit'])){ $FirstName = mysql_real_escape_string($_POST["FirstName"]); $LastName = mysql_real_escape_string($_POST["LastName"]); $DeptName = mysql_real_escape_string($_POST["DeptName"]); $AssetCategory = mysql_real_escape_string($_POST["AssetCategory"]); $Model = mysql_real_escape_string($_POST["Model"]); $Maker = mysql_real_escape_string($_POST["Maker"]); $SerialNum = mysql_real_escape_string($_POST["SerialNum"]); $DateAguired = mysql_real_escape_string($_POST["DateAguired"]); $AssetDescription = mysql_real_escape_string($_POST["AssetDescription"]); $AssetCatID = mysql_real_escape_string($_POST["AssetCatID"]); $AssetID = mysql_real_escape_string($_POST["AssetID"]); $EmpID = mysql_real_escape_string($_POST["EmpID"]); $DeptID = mysql_real_escape_string($_POST["DeptID"]); if(empty($FirstName) || empty($LastName) || empty($DeptName) || empty($AssetCategory) || empty($Model) || empty($Maker) || empty($SerialNum) || empty($DateAguired) || empty($AssetDescription)) { print "Please feel in all fields"; } else { ///insert into the department table.......... $query1= "INSERT INTO department VALUES (null,'{$DeptName}')"; $result1 = mysql_query($query1) or die(mysql_error()); $DeptID = mysql_insert_id(); //insert into the assetcategory table $query2= "INSERT INTO assetcategory VALUES (null,'{$AssetCategory}')"; $result2 = mysql_query($query2) or die(mysql_error()); $AssetCatID = mysql_insert_id(); //insert into the assets table $query3= "INSERT INTO assets VALUES (null,'{$AssetDescription}', '{$EmpID}','{$AssetCatID}','{$DeptID}','{$Model}','{$Maker}','{$SerialNum}','{$DateAguired}')"; $result3 = mysql_query($query3) or die(mysql_error()); $AssetCatID = mysql_insert_id(); print $query1; } } else { ?> the page is displaying this error: "Column 'AssetCatID' cannot be null" I would need help in inserting into all table required data. any help? hope i am making sense Hi, I'm trying to insert data into two different tables using, but am getting an error I can't figure out. If I move the $mysqli->commit(); into the foreach loop, I get at least one returned row before the rest fail. The error current error message is Array ( [0] => Error: Couldn't insert into english! ). Any idea what is causing this?
<?php
$file_array = file('../grammar/conjunctions.txt');
$csv = array_map('str_getcsv', $file_array);
// DB
$mysqli = new mysqli('localhost', 'root', '******', 'angos');
$mysqli->autocommit(false);
$error = array();
foreach($csv as $value)
{
$angos_query = $mysqli->query("INSERT INTO angos (angos, grammar) VALUES ('$value[0]', 'con')");
$id = $mysqli->insert_id; // grab the currant angos table id
if($angos_query == false)
{
array_push($error, "Error: Couldn't insert into angos!");
}
$english_query = $mysqli->query("INSERT INTO english (angos_id, english) VALUES ('$id', '$value[1]')");
if($english_query == false)
{
array_push($error, "Error: Couldn't insert into english!");
}
if(!empty($error))
{
$mysqli->rollback();
}
}
$mysqli->commit();
print_r($error);
// print_r($csv);
?>
More info SQL:
CREATE TABLE angos
(
id int unsigned NOT NULL AUTO_INCREMENT PRIMARY KEY,
angos varchar(255) not null,
grammar varchar(3) not null,
updated TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
CONSTRAINT unique_input UNIQUE (angos)
) engine=InnoDB;
CREATE TABLE english
(
id int unsigned not null primary key,
angos_id int unsigned,
english varchar(255),
grammar_note varchar(500),
CONSTRAINT fk_angos_source FOREIGN KEY (angos_id) REFERENCES angos(id) ON DELETE CASCADE ON UPDATE CASCADE
) engine=InnoDB;
$username = $_POST['uid'];
$email = $_POST['mail'];
$password = $_POST['pwd'];
$passwordRepeat = $_POST['pwd-repeat'];
$date = $_POST['date2'];
$stream = $_POST['relationship'];
$sql1 = "INSERT INTO users (uidUsers, emailUsers, pwdUsers, relationship) VALUES (?, ?, ?, ?);";
$sql2 = "INSERT INTO Family1 (username, application_filed, relationship) VALUES (?, ?, ?);";
$sql3 = "INSERT INTO Family2 (username, application_filed, relationship) VALUES (?, ?, ?);";
mysqli_query($sql1, $conn);
mysqli_query($sql2, $conn);
mysqli_query($sql3, $conn);
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql2)) {
header("Location: ../signup.php?error=sqlerror");
exit();
}
else {
mysqli_stmt_bind_param($stmt, "sss", $username, $date, $stream);
$result = mysqli_stmt_get_result($stmt);
if ($row = mysqli_fetch_assoc($result))
($username==$_SESSION['uid'] and $stream =='nursing');
mysqli_stmt_execute($stmt);
}
if (!mysqli_stmt_prepare($stmt, $sql3)) {
header("Location: ../signup.php?error=sqlerror");
exit();
}
else {
mysqli_stmt_bind_param($stmt, "sss", $username, $date, $stream);
$result = mysqli_stmt_get_result($stmt);
if ($row = mysqli_fetch_assoc($result))
($username==$_SESSION['uid'] and $stream =='doctoral');
mysqli_stmt_execute($stmt);
}
if (!mysqli_stmt_prepare($stmt, $sql1)) {
header("Location: ../signup.php?error=sqlerror");
exit();
}
if (!mysqli_stmt_prepare($stmt, $sql1)) {
header("Location: ../signup.php?error=sqlerror");
exit();
}
else {
$hashedPwd = password_hash($password, PASSWORD_DEFAULT);
mysqli_stmt_bind_param($stmt, "ssss", $username, $email, $hashedPwd, $stream);
mysqli_stmt_execute($stmt);
header("Location: ../signup.php?signup=success");
exit();
I was wondering if someone could point me in the right direction. I have this code. They idea I had behind it is to insert values into different tables depending on variables being passed. So when user fills out a form and selects $stream="nursing" I want results to go to table 'users' and 'Family1', but not 'Family2' table. and if user selects $stream='doctoral' results should go to table 'users' and 'Family2', and not go to 'Family1' But with my query I get results go to both table and also users table. And there is no restriction to what users selects, variable $stream being passed no matter what it is. Is this the wrong way to go here? Did I completely mess up the logic? Here is my java script
$.each(data, function() {
var myTable1 =
'<tr bgcolor=#ffffff>' +
'<td align=center>'+this.vehiclelog_plate+'</td>' +
'<td align=center>'+this.vehiclelog_name+'</td>' +
'<td align=center>'+this.vehiclelog_date+'</td>' +
'<td align=center>'+this.vehiclelog_reftype+'</td>' +
'<td align=center>'+this.vehiclelog_refid+'</td>' +
'<td align=center>'+this.vehiclelog_description+'</td>' +
'</tr>';
$('#tableField').append(myTable1);
});
And here is my HTML
<div id="tabs-6">
<div id="rform">
<form action = "home.php" method="post">
<fieldset>
<legend>View Vehicle Service</legend><br><br>
<div>
<label class="label-left">Vehicle Name:</label>
<span role="status" aria-live="polite" class="ui-helper-hidden-accessible"></span>
<input id="autocomplete" title="type "a"" class="ui-autocomplete-input" name="vehicle_service_search" autocomplete="off">
</div>
<div>
<button id="searchVehicleDesc" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" role="button" aria-disabled="false">
<span class="ui-button-text">View</span>
</button>
</div>
<div>
<table border="1" width="650" id = "tableField">
<tr bgcolor=#c0c0c0>
<td width=100 align="center"><font face="helvetica"><b>Vehicle Name</b></font></td>
<td width=80 align="center"><font face="helvetica"><b>Service Type</b></font></td>
<td width=80 align="center"><font face="helvetica"><b>Others</b></font></td>
<td width=100 align="center"><font face="helvetica"><b>Reference No.</b></font></td>
<td width=80 align="center"><font face="helvetica"><b>Reference Date</b></font></td>
<td width=80 align="center"><font face="helvetica"><b>Reference Type</b></font></td>
</tr>
</table>
</div>
</fieldset>
</form>
</div>
</div>
And here is my response
[{"success":1,"plate":"SIX666","name":"YAMAHA ISUZU","date":"0000-00-00","type":"Add Vehicle","id":"0","desc":""},{"success":1,"plate":"VIE597","name":"Resource id #6","date":"2014-05-09","type":"OIL","id":"332142","desc":"castroil"}]
Im getting the right response from ajax but I can't put my value to a table I get an alert that says undefined.
So I have a database that is structured like this: https://imgur.com/a/DdyTqiE
I would like to loop through this table, and get a count of how many were 'is_no_show' and 'is_cancelled' per (unique) category with respect to 'scheduled_student_services.' How would I approach this? I know I probably need to do two loops but I'm not sure the PHP syntax for this. Any suggestions or tips would be helpful. Thank you for your time! How would you retrieve information from one table and enter that informatio to another and the same time get other details from form. Do you retrieve the data and use a form to post this and insert that to another table. Have I confused you??? Hello, I'm new! I am trying to populate a list of locations based on ratings. I can populate the list just fine but it's displaying every record in the table. Is there a way of only looping 4 times so as to just display the top 4 records? Here's the code: Code: [Select] $result = mysql_query("SELECT * FROM locations ORDER BY rating DESC"); while($row = mysql_fetch_array($result)) { echo $row['name']; echo "<br>; } Thankyou in advance for any help Hi there reader(s) I've been searching everywhere, and I'm not one to post my issue unless I've searched as hard as I could, I'm not new to PHP, but I am new to MySQL (so go easy on me) and I've got variables all set up, and I've successfully established connection to my Database and my Database's table, but now I cannot find a way to insert the data (variable) into a column of the table. Willing to provide more information! I want to allow users to post entries to a NEWS table and, if they wish, to post an accompanying image to an IMAGES table. Tables are like this: NEWS id // if there'll be an accompanying image, this id to be sent to IMAGES table title subtitle created news_entry category IMAGES image_id f_news_id // the foreign id of the associated post in NEWS table filename caption description So, the user comes to the insert_entry.php page and creates a post. If the user clicks an "Upload accompanying image" link, the news post id must be inserted in the f_news_id field when the image is uploaded. Code excerpt from insert_entry.php: Code: [Select] // Insert the news_entry in the database... // Make the query: $q = "INSERT INTO news (title, subtitle, news_entry, category) VALUES ('$title', '$subtitle', '$news_entry', '$category') "; $r = @mysqli_query ($dbc, $q); // Run the query. if ($r) { // If it ran OK. // Print a message: echo "<h1>Thank you!</h1> <p>You have successfully inserted the News Entry below.</p>"; echo "<h1>" . stripslashes($title) . "</h1><h2>" . stripslashes($subtitle) . "</h2><p>" . stripslashes($news_entry) . "</p>"; // get id of record just created $q = "SELECT id FROM news ORDER BY created DESC LIMIT 1"; $r = mysqli_query($dbc, $q); while ($row = mysqli_fetch_assoc($r)) { // pass the id via GET in the URL echo "<a href='upload_image.php?=" . $row['id'] . "'>Upload image</a>"; } mysqli_close($dbc); ?> Code excerpt from upload_image.php: Code: [Select] // insert news post id into images table if user came via insert_entry.php page // Make the query: require_once ('includes/mysqli_connect.php'); // Connect to the db. $description = mysqli_real_escape_string($dbc, trim($_POST['description'])); $caption = mysqli_real_escape_string($dbc, trim($_POST['caption'])); if (isset($_GET['id'])) { // if there's a NEWS post id $q = "INSERT INTO images (f_news_id, filename, caption, description) VALUES ('$_GET['id']', '{$_FILES['upload']['name']}', '$caption', '$description')"; } else { // if user arrived at upload_image.php otherwise and there's *not* a NEWS post id $q = "INSERT INTO images (filename, caption, description) VALUES ('{$_FILES['upload']['name']}', '$caption', '$description') "; } $r = @mysqli_query ($dbc, $q); // Run the query. if ($r) { // If it ran OK. // Print a message: echo "<p>Info entered in images table.</p>"; Am I going about this the wrong way? Am new to php... so any advice much appreciated... im trying to write a script takes an xml files with tv show info, splits it into the show and the eppisode info and the place it into a table, i have got it to proccess all the info and print it out on a web page, but i cant, for the life of me, get it to insert said data into the table, it seems to be just ignoring the code and prints out the data as if nothing happens, no errors or anything. here is my code (abit messy but im only just starting and its my test.php) Code: [Select] <?php $tvdb_mirror = "http://www.thetvdb.com/api/"; $tvdb_time = "http://www.thetvdb.com/api/Updates.php?type=none"; $dbname = "mediadb"; $dbuser = "root"; $dbpass = ""; $dbserv = "127.0.0.1"; $rss = simplexml_load_file('sample.xml'); $showName = "Show Name = ".$rss->Series->SeriesName; print $showName; print "<br />Show Discription = ".$rss->Series->Overview; print "<br />"; mysql_connect('127.0.0.1', 'root', ''); @mysql_select_db('mediadb') or die("Unable to select database"); foreach ($rss->Episode as $item) { $seasonnum = $item->Combined_season; $EpisodeNumber = $item->EpisodeNumber; if($EpisodeNumber < 10){ $EpisodeNumber = "0".$EpisodeNumber; }; $EpisodeName = $item->EpisodeName; $Overview = $item->Overview; $airdate = $item->FirstAired; $tvdbid = $item ->id; $query = "INSERT INTO eppisodes VALUES('', '1', ".$EpisodeName.", ".$Overview.", ".$airdate.", '1', ".$tvdbid.", '-1', ".$seasonnum.", ".$EpisodeNumber.")"; mysql_query($query); print "<br />".$showName." - ".$seasonnum."x".$EpisodeNumber." - ".$EpisodeName." Overview:<br />".$Overview; } mysql_close(); ?> i am a noob @ php and mysql, but i have doubke and triple checked the names of the db and table. here is an sql dump of my db Code: [Select] -- phpMyAdmin SQL Dump -- version 3.3.5 -- http://www.phpmyadmin.net -- -- Host: 127.0.0.1 -- Generation Time: Nov 13, 2010 at 12:48 PM -- Server version: 5.1.49 -- PHP Version: 5.3.3 SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO"; /*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */; /*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */; /*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */; /*!40101 SET NAMES utf8 */; -- -- Database: `mediadb` -- -- -------------------------------------------------------- -- -- Table structure for table `eppisodes` -- CREATE TABLE IF NOT EXISTS `eppisodes` ( `id` int(11) NOT NULL AUTO_INCREMENT, `showID` int(11) NOT NULL, `eppname` varchar(255) NOT NULL, `eppdesc` longtext NOT NULL, `airdate` date NOT NULL, `format` int(11) NOT NULL, `tvdbid` varchar(20) NOT NULL, `dohave` tinyint(1) NOT NULL, `season` varchar(2) NOT NULL, `eppisode` varchar(3) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; -- -- Dumping data for table `eppisodes` -- -- -------------------------------------------------------- -- -- Table structure for table `shows` -- CREATE TABLE IF NOT EXISTS `shows` ( `id` int(100) NOT NULL AUTO_INCREMENT, `name` varchar(255) NOT NULL, `description` longtext NOT NULL, `TVDBID` int(100) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; -- -- Dumping data for table `shows` -- INSERT INTO `shows` (`id`, `name`, `description`, `TVDBID`) VALUES (1, 'higogo', 'some info', 67546); any help would be very much appericiated. fyi im running win7 with easyPHP 5.3.3 with php 5.3.3, mysql 5.1.49 apache 2.2.16 Stuck again on simple code. I am trying to insert some fields extracted from one table into another. I'm using code that worked elsewhere. The SQL statement flies, the script runs, the input array is printed back I get an echo back from the end of the script but nothing is added to the table. Even aded an echo print_r in the conditional and I know the data is getting to the execute command. The script follows with a sample of the input array. I have attached am image of the table I am trying to insert the data into. --Kenoli
The script:
<?php
require '__classes/DB.php';
$sql = "SELECT name, table_id, image_name, description, medium FROM tbl_person_data ";
$stmt = $pdo->query($sql);
$array1 = $stmt->fetchall(PDO::FETCH_ASSOC);
$stmt = $pdo->prepare("INSERT INTO Images (name, person_id, filename, description, medium) VALUES (?,?,?,?,?)");
//$pdo->beginTransaction();
foreach ($array1 as $row)
{
$stmt->execute($row);
}
echo "<pre>";
print_r ($row);
echo "</pre>";
echo '<h4>Got to end of file</h4>';
?>
$array1: The input array
[0] => Array
(
[name] => Carol Lettko
[table_id] => 21
[image_name] => Carol_Lettko-DSC_3022.jpg
[description] => Baby Herons/Brickyard
[medium] => photo
)
[1] => Array
(
[name] =>
[table_id] => 22
[image_name] => Carol_Lettko-DSC_0164.JPG
[description] => Heron/Brickyard
[medium] => photo
)
[2] => Array
(
[name] =>
[table_id] => 23
[image_name] => Carol_Lettko-IMG_5723.jpg
[description] => Kayaker/Brickyard
[medium] => photo
)
i want insert data from text box in html form.there are many text boxes gave name using 2 loops..please tell how can insert data to table?please reply 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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