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PHP - Allowing Foreign Letters?
Would someone tell me how to allow foreign letters, normal english letters and numbers using the preg_replace php function?
Similar TutorialsHello, I posted here as although it's in regards to mysql the thing i want to ask is in regards to php itself. I have two tables below as follows: Code: [Select] CREATE TABLE `users` ( `uid` int(11) NOT NULL AUTO_INCREMENT, `status` char(10) NOT NULL, `username` varchar(15) NOT NULL, `email` varchar(50) NOT NULL, `password` char(32) NOT NULL, `reg_date` int(11) NOT NULL, `ip` varchar(39) DEFAULT NULL, PRIMARY KEY (`uid`), UNIQUE KEY `username` (`username`,`email`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; CREATE TABLE `users_profiles` ( `uid` int(11) NOT NULL, `first_name` varchar(40) DEFAULT NULL, `last_name` varchar(50) DEFAULT NULL, `gender` char(6) DEFAULT NULL, `website` varchar(100) DEFAULT NULL, `msn` varchar(60) DEFAULT NULL, `aim` varchar(60) DEFAULT NULL, `yim` varchar(60) DEFAULT NULL, `twitter` varchar(15) DEFAULT NULL, UNIQUE KEY `uid` (`uid`), CONSTRAINT `users_profiles_ibfk_1` FOREIGN KEY (`uid`) REFERENCES `users` (`uid`) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=utf8; From the above you can see the uid in the users_profiles table is a foreign key and references the uid in the users table to link both tables together. My scenario is this: A user fills out a registration form that asks for username, email and a password. After validating the form data etc it is inserted into the users table. Now at this stage they have no profile in the users_profiles table as they only just signed up. They login and xxx data is stored in a session along with there uid from users table. They now visit there profile, they fill in the profile form fields and submit the form. Now can someone tell me if what i think i should be doing is correct, if not please tell me what i should do or a better way: After validating the data that was submitted on profile form etc i could do something like a simple if else statement which will first check to see if the uid in session which is from users table matched a uid in the users_profiles table. If there's no match which in this case it won't be as they are a new user and don't have a profile record in users_profiles table then do an INSERT query into users_profiles table which would insert the submitted values from the form and the uid from session, on success that would now link the user from users table to there newly created profile in the users_profiles table. But if the uid in session matched a uid in user_profiles then obviously this means the user already has a profile ie.. submitted profile info previously so do an UPDATE query instead. Am i rite in thinking this is how i would do it? example below, note it's not valid code just to try an illustrate what i am saying; Hopefully! Code: [Select] <?php if($_SESSION['uid'] == user_profiles.uid) { /* If the uid in session which is the uid from users table * (we got it on successful login) * matches a users uid in user_profiles table then profile already exists * for user so do an UPDATE query here */ } else { /* $_SESSION['uid'] does not match a uid in users_profiles table * so instead do a INSERT QUERY here. This means no profile exists for user. * The uid from session is inserted into users_profiles table (column uid) * which links the profile to user. */ } ?> I know if i delete a user it will also delete there profile if they have one like it should do to obviously not leave redundant data in the database and to ensure data intergity (sorry i think that's what it's called) Thanks for any help. PHPFAN Hi, I'm having following problem when I try to add records in two tables, one (password) has primary key and student has a foreign key contraint with password. Anyone please clarify what Im doing wrong. Thanks ! SQL CODE: INSERT INTO student(id, firstname, surname,dob, class,nationality,houseno,line1,line2,city) VALUES('00010', 'test','test','1980-01-01','test','test','test','test','test','test'); INSERT INTO password (id, password,salt) VALUES('00010','abc','abcde'); ERROR: #1452 - Cannot add or update a child row: a foreign key constraint fails (`chr_fyp/student`, CONSTRAINT `student_ibfk_1` FOREIGN KEY (`id`) REFERENCES `password` (`id`) ON DELETE CASCADE ON UPDATE CASCADE) I've hit a bit of a roadblock with one of my projects and I cant seem to get past it. I am trying to add a new row to one of my tables, but the foreign key seems to be causing me problems. As you will see from my code below, I am trying to insert a row into my 'job' table. Error: Query failed: Cannot add or update a child row: a foreign key constraint fails (`kanix_support_desk`.`job`, CONSTRAINT `job_ibfk_1` FOREIGN KEY (`agentId`) REFERENCES `agents` (`agentId`)) Tables: Code: [Select] CREATE TABLE IF NOT EXISTS `agents` ( `agentId` int(11) NOT NULL AUTO_INCREMENT, `AgentFirstname` varchar(50) DEFAULT NULL, `AgentSurname` varchar(50) DEFAULT NULL, `username` varchar(25) DEFAULT NULL, `password` varchar(25) DEFAULT NULL, `rank` varchar(12) NOT NULL, PRIMARY KEY (`agentId`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; CREATE TABLE IF NOT EXISTS `job` ( `jobId` int(11) NOT NULL AUTO_INCREMENT, `agentId` int(11) NOT NULL, `date` date NOT NULL, `time` time DEFAULT NULL, `jobName` varchar(100) NOT NULL, `jobDescription` varchar(9999) NOT NULL, `jobstatus` varchar(50) NOT NULL, `customerFirstname` varchar(50) DEFAULT NULL, `customerSurname` varchar(50) DEFAULT NULL, `CustomerTelephoneNo` varchar(50) DEFAULT NULL, PRIMARY KEY (`jobId`), KEY `agentId` (`agentId`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; Php: Code: [Select] session_start(); include("dbConnect.php"); $agent = $_SESSION['agent']; //Set agent name and ID variables $agentid = $_SESSION['identity']; //Check that no section of the form was left blank if($_POST['fname'] != null && $_POST['sname'] != null && $_POST['phone'] != null && $_POST['jtitle'] != null && $_POST['jdesc'] != null){ mysql_query("INSERT INTO job (jobId, agentId, date, time, jobName, jobDescription, jobstatus, customerFirstname, customerSurname, customerTelephoneNo) VALUES (null, '".$agentid."', CURDATE(), NOW(), '".$_POST['jtitle']."', '".$_POST['jdesc']."', 'Live', '".$_POST['fname']."', '".$_POST['sname']."', '".$_POST['phone']."')") or die ("Query failed: " . mysql_error()); header("location:reception.php?content=3"); } else{ //Error message header("location:reception.php?content=4"); } 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... How do I convert accented characters etc into their code form so I can put them into my database? I have tried everything I can think of like htmlentities, htmlspecialchars, url_decode etc and nothing seems to work! I am storing content in a table with different fields for different languages, e.g. polish, spanish etc. I can use the following function: function getRewriteString($sString) { $string = strtolower(htmlentities($sString)); $string = preg_replace("/&(.)(uml);/", "$1e", $string); $string = preg_replace("/&(.)(acute|cedil|circ|ring|tilde|uml);/", "$1", $string); $string = preg_replace("/([^a-z0-9]+)/", " ", html_entity_decode($string)); $string = trim($string, "-"); return $string; } Which replaces accented chars with their unaccented counterparts but it doesn't work 100% of the time and isn't really what I want because I'd like to have the accented chars as they will be used on different language sites. Thanks in advance. I have two tables in my MySQL database: 1. user user_id | firstname | lastname | nickname etc. and 2. con (for contribution) con_id | user_id (foreign key) | name | contribution | category etc. Here's how I wanted to solve the problem of assigning the foreign key to the 2nd table (I'm a beginner so bear with me. ) When logging in, I assigned the user_id to the session variable like this: // do the while loop while ($assoc = mysqli_fetch_assoc($rows)) { // assign database COLUMNS to the variables $dbuser_id = $assoc['user_id']; $dbuser_name = $assoc['nickname']; $dbuser_password = $assoc['password']; etc..................... // set a session after login $_SESSION['user'] = $dbuser_name . $dbuser_id; This prints out the user_id just like I wanted to: echo "Your user_id is: " . $_SESSION['user_id'] = $dbuser_id; Gives: Code: [Select] Your user_id is: 35 ... and this is how the assignment of the foreign key looks like WHILE posting the contribution: $user_id = $_SESSION['user'] = $dbuser_id; if (!empty($knuffix_name) && !empty($knuffix_category) && !empty($knuffix_contribution)) { // Write the data into the database $query = sprintf("INSERT INTO con (con_id, user_id, name, contribution, category, contributed_date) VALUES (' ', '$user_id', '%s', '%s', '%s', now())", mysqli_real_escape_string($dbc, $knuffix_name), mysqli_real_escape_string($dbc, $knuffix_contribution), mysqli_real_escape_string($dbc, $knuffix_category)); When I now do a contribution through the input areas, nothing gets inserted into the database and I automatically get logged out. Obviously it's not working as I thought it would and I'd like to know why is that? Notice that if I take the user_id part OUT, it's working again, so the rest of the code must be right then. Another question I have is: Is this common practice to solve this problem of assigning a foreign key, or is there a better way of doing it? Hi, I cannot find where to add foreign key constraints in phpmyadmin! Any help is appreciated! I have a table that holds a set of matches in a football (soccer) league. On a teams individual page I want to show just their matches and am very close to getting it right but just cannot get there! I am using 2 tables for this.... teams team_id team_name 1 TEAM1 2 TEAM2 3 TEAM3 4 TEAM4 and all_games game_id home_team home_score away_team away_score 1 1 2 2 0 2 3 0 4 1 3 4 2 1 1 4 2 0 3 0 Both 'home_team' and 'away_team' are foreign keys to teams.team_id. The query that I have used to almost get it working is SELECT * FROM all_games, teams WHERE '%s' IN (home_team, away_team ) AND all_games.home_team=teams.team_id And the html is Code: [Select] <tr> <td>all_games_id</td> <td>home_team</td> <td>home_goals</td> <td>away_team</td> <td>away_goals</td> </tr> <tr> <td><?php echo $row_all_games['all_games_id']; ?></td> <td><?php echo $row_all_games['team_name']; ?></td> <td><?php echo $row_all_games['home_goals']; ?></td> <td><?php echo $row_all_games['team_name']; ?></td> <td><?php echo $row_all_games['away_goals']; ?></td> </tr> This pulls the correct games into the page but will show the 'home_team.team_name' as both 'home_team' and 'away_team'. For example the page for TEAM1 shows up as match_id home_team home_score away_team away_score 1 TEAM1 2 TEAM1 0 3 TEAM4 2 TEAM4 1 What it should show is match_id home_team home_score away_team away_score 1 TEAM1 2 TEAM2 0 3 TEAM4 2 TEAM1 1 Obviously in the current query there is nothing to show that the 'away_team' is also equal to 'teams.team_id'. I have tried this query SELECT * FROM all_games, teams WHERE '%s' IN (home_team, away_team ) AND (all_games.home_team=teams.team_id OR all_games.away_team=teams.team_id) But this gives the exact opposite result with just the 'away_team.team_name' showing up as 'home_team' and 'away_team' match_id home_team home_score away_team away_score 1 TEAM2 2 TEAM2 0 3 TEAM1 2 TEAM1 1 while SELECT * FROM all_games, teams WHERE '%s' IN (home_team, away_team ) AND (all_games.home_team=teams.team_id AND all_games.away_team=teams.team_id) throws up an empty query. Does anyone have any suggestions on how this can be done? I have exhausted my patience with Google simply as I do not know what the correct search term is. As an aside, when I do get it working, I want the format on the page to be slightly different in that instead of showing both home_team and away_team names I would want it to show "H" if the team in question is the 'home_team' and "A" if it is the 'away_ team' plus the name of the team that they played and whether the 'home_score' or 'away_score' belonged to them, as well as whether they won, drew or lost the game. So, staying with the page for TEAM1 it would show as MATCH ID HOME or AWAY OPPONENT W/D/L SCORE FOR SCORE AGAINST 1 H TEAM2 Won 2 0 3 A TEAM4 Lost 1 2 I am currently planning on doing this with IF statements in the html, something along these lines if home_team = %s echo "Home" else echo "Away" and if home_team = %s AND home_score>away_score echo "Won" else if away_team = %s and away_score>home_score echo "Won" else if home_score= away_score echo "Drew" else echo "Lost If anyone can think of a better way to do this ( I am naturally assuming that it would not work in the way I intend!)I would be extremely greatful too! Thanks in advance for any advice, pointers or offers to write the whole code!!! Steve For example: Code: [Select] DELETE FROM user_question_answer WHERE user_id = 1 AND question_id = 7 AND `answer_id` = 3 This sql query has no error, but reports deleting 0 rows. user_id and question_id fields are both foreign keys in this table. Could that be why it's not working, yet not reporting an error? Thank you for your time. This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=359282.0 SET UP: Windows vista # XAMPP 1.7.3, # Apache 2.2.14 (IPv6 enabled) + OpenSSL 0.9.8l # MySQL 5.1.41 + PBXT engine # PHP 5.3.1 # phpMyAdmin Problem: The name_id column from the card_numbers table doesn't increment at all. Even if if I put in two rows of data I'm left with the name_id not increasing. Main Goal: I want that column to increment with the same values as the other tables. Code for Mysql database: <?php require 'db.inc.php'; $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); $query = 'CREATE TABLE IF NOT EXISTS subscriptions ( name_id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, name VARCHAR(50) NOT NULL, email_address VARCHAR(50), membership_type VARCHAR(50), PRIMARY KEY (name_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); // create the user information table $query = 'CREATE TABLE IF NOT EXISTS site_user_info ( name_id INTEGER UNSIGNED NOT NULL, terms_and_conditions VARCHAR(50) NOT NULL, name_on_card VARCHAR(50), credit_card_number VARCHAR(50), FOREIGN KEY (name_id) REFERENCES subscriptions(name_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); $query = 'CREATE TABLE IF NOT EXISTS card_numbers ( name_id INTEGER UNSIGNED NOT NULL, credit_card_expiration_data VARCHAR(50), FOREIGN KEY (name_id) REFERENCES subscriptions(name_id) ) ENGINE=MyISAM'; mysql_query($query, $db) or die (mysql_error($db)); echo 'Success!'; ?> Code of the data being inserted into mysql from forms: <?php //let's start our session, so we have access to stored data session_start(); session_register('membership_type'); session_register('terms_and_conditions'); include 'db.inc.php'; $db = mysql_connect('localhost', 'root', '') or die ('Unable to connect. Check your connection parameters.'); mysql_select_db('ourgallery', $db) or die(mysql_error($db)); //let's create the query $query = sprintf("INSERT INTO subscriptions ( name_id, name, email_address, membership_type) VALUES ('%s','%s','%s','%s')", name_id, mysql_real_escape_string($_SESSION['name']), mysql_real_escape_string($_SESSION['email_address']), mysql_real_escape_string($_SESSION['membership_type'])); //let's run the query $result = mysql_query($query, $db) or die(mysql_error($db)); $name_id = mysql_insert_id(); $query = sprintf("INSERT INTO site_user_info ( name_id, terms_and_conditions, name_on_card, credit_card_number ) VALUES ('%s','%s','%s','%s')", mysql_insert_id(), mysql_real_escape_string($_SESSION['terms_and_conditions']), mysql_real_escape_string($_POST['name_on_card']), mysql_real_escape_string($_POST['credit_card_number'])); //let's run the query $result = mysql_query($query, $db) or die(mysql_error($db)); $name_id = mysql_insert_id(); $query = sprintf("INSERT INTO card_numbers ( name_id, credit_card_expiration_data) VALUES ('%s','%s')", mysql_insert_id(), mysql_real_escape_string($_POST['credit_card_expiration_data'])); $result = mysql_query($query, $db) or die(mysql_error($db)); echo '$result'; ?> Should I be JOINING these tables together? I don't want the IDs to get muddled up otherwise it will screw up my database. I've set up a staging server with Ubuntu 14.04, Apache, PHP, and MySQL, and I'm having a configuration issue that I need to resolve.
index.html uses a server side include. I was required to change the shtml file it references to a php file, but it's not working properly.
When my webpage uses the code:
<?php include ("footer.shtml"); ?>
then the page loads correctly
However, when I change it to:
<?php include ("footer.php"); ?>
then I instead see a message stating,
"This attachment was removed because it contains data that could pose a security risk."
I can't determine which configuration file I need to change, and what the change needs to be.
I was wondering if there was a way to allow users to input there own HTML and CSS in like a members only site using PHP? If so, how would it be done? here is my form layout, Title: txtfield URL: txtfield Category: dropdown (pulls the just category fields in my "categories" table submit - when working right this form is suppose to enter this info into my "movies" table. Code to form.php Code: [Select] <html> <form id="form1" name="Update" method="post" action="add3.php"> <label> Title: <input type="text" name="Title" id="Title" /> </label> <br /> <label> URL: <input type="text" name="URL" id="URL" /> </label> <select name='dropdown' id='dropdown'> <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("mydb", $con); $query = "SELECT Category from categories"; $result = mysql_query($query) OR DIE ("There was an error" .mysql_error()); while($row=mysql_fetch_array($result)) { $category = $row['Category']; echo " <option value=\"$category\">$category</option>"; } php?> </select> <input name="" type="submit" value="send" /> </form> </html> then here is the code to the process page (add.php) Code: [Select] <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("m5", $con); $sql="INSERT INTO movies (Category, URL, Title) VALUES ('$_POST[dropdown]','$_POST[URL]','$_POST[Title]')"; if (!mysql_query($sql,$con)) { die ('Error: ' . mysql_error()); } echo "Record added"; after i submit my form i get "Error: Cannot add or update a child row: a foreign key constraint fails (`mydb`.`movies`, CONSTRAINT `movies_ibfk_1` FOREIGN KEY (`Category`) REFERENCES `categories` (`ID`) ON DELETE CASCADE ON UPDATE CASCADE)" correct me if im wrong but im almost %99 sure that my tables are set up correctly for my project. just to be sure, movies ------------ ID (PK) Title Category (FK) URL categories --------------- ID (PK) Category at first i was also getting a "undefined index" along with this error but i reversed the values in the INSERT so now im just getting the CONSTRAINT error so this now leaves me to beleive that i may have made a error one of the tags possibly. also while i was looking around i decided to create a different procces page hoping i would run into some quick luck: add3.php Code: [Select] <?php $Category=$_POST['dropdown']; $URL=$_POST['URL']; $Title=$_POST['Title']; $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("mymb", $con); $query = "INSERT INTO movies VALUES ('$Category','$URL','Title')"; $result= mysql_query($query); if ($result) { echo "Successful"; } else { echo "Failed"; } mysql_close(); ?> and this one gives me "Failed" error. i have noticed that there are many many ways of doing this so i went ahead and posted both process page hoping this could give some better ideas rather then confusion (i hope anyways). im not really sure which ones its more along the lines of php/form "standard" if there is any but whichever may be the easiest for you to take a look at it doesnt matter and i would be greatly thankful. IMO i think its just a varriation of not having the formats right and just not taylored correctly to my needs. many thanks in advanced. I have a bit of code (maybe a lot, i dunno) that works pretty well. However, Im having some issues. Im using file_get_contents() to get a remote page from another server, and then displaying it on the current page. The issue? the images/css/urls are relative to the other domain, and so they don't display. I know httrack uses something like this, but it doesn't show the webpage on the server, it downloads the page. my thoughts are that if i could use some php function to change all urls to not be relative, it would work, but i cant do that. anyway, here is my code, you can try it for yourself too, if you want to to see what i mean. tryit: http://dageek247.tk/viewit/ code: <html> <head> <style type="text/css"> .perc {width:93%; height:90%;} .perk {width:6%; height:90%;} </style> </head> <?PHP if ($_GET['url'] == "") { $url = 'http://dageek247.tk/wordpress/'; } else { $url = $_GET['url']; } ?> <body> <div style="height:25px;width:100%;"> <div id="floatbar" style="background:lightblue; width:100%; height:25px; position:fixed; left:0;top:0; z-index:10;"> <center> <form method='get' action='index.php'> <input type='url' name='url' value='<?PHP echo $url; ?>' class='perc'> <input type='submit' value='Go!' class='perk'> </form> </center> </div> </div> <div id='showview' style='width:100%; height:100%;'> <?php //download the webpage $contents = file_get_contents($url); /* //mess around with the webpages code $contints = explode("\n", $contents); //go through each line and cahnge src and href code foreach( $contints as $key => $value){ //images $eqsrc = "src=\""; $src = strpos($original, $eqsrc); $src_end = $src + strlen($eqsrc); $new_src = substr_replace($value, $eqsrc & $url, $src_end, 0); //links $eqhref = "href=\""; $href = strpos($original, $eqhref); $href_end = $href + strlen($eqhref); $new_href = substr_replace($value, $eqhref & $url, $href_end, 0); } $contents = implode("\n", $contints); */ //attempt to be able to download images this way $_SERVER['HTTP_HOST'] = $url; //fails :( //view the webpage echo $contents; ?> </div> </body> </html> You can also see how i have tried to change relative urls into absolute ones in the commented out area. it doesnt work. Hey, I've created a messaging system for my site and I'm trying to make it so that a copy of the original message is included and automatically put into the message content box when somebody clicks to reply. I've done this literally just by making the "reply" button into a form which also posts the content of the message from the previous page and puts it into the content box. However I have some issues with it: firstly, wherever ' is used, it becomes \' and wherever " is used, the whole of the rest of the message is blanked out. Can anybody point me in the direction of the solutions I might need to make it so that ' and " both appear normally without cutting the rest of the message off? Thanks in advance & let me know if anything is missing! Hi,
I want each new incident created into the t_incidents table by the user to be associated with the AgencyID, a foreign key in the t_users table.
The problem is I do not know how this relationship will work and whether I will need a junction table.
The two tables a
t_Users +----------+----------+------------+ |UserID | AgencyID |User name | +----------+----------+------------+ |1 | 1 |john | +----------+----------+------------+ |2 | 1 |andrew | +----------+----------+------------+ t_Agencies +----------+------------+ |AgencyID |agency name | +----------+------------+ |1 |police | +----------+------------+ |2 |immigration | +----------+------------+I will appreciate your advice. Joseph Edited by josephbupe, 13 October 2014 - 03:04 AM. I have a table that contains the schools in my system: schools id name location ... Then, I have three tables that use the id of this table: schoolAdmins schoolID schoolContests schoolID students schoolID When I go to delete a school from my system, I want to check to see if that school is connected to any of these three other tables first. This is what I tried (but obviously failed because I'm here) where I'm passing the query the $studentID in question: SELECT * FROM schoolAdmins, schoolContests, students WHERE (schoolAdmins.schoolID = $schoolID) OR (schoolContests.schoolID = $schoolID) OR (students.schoolID = $schoolID) I'm really new to the concept of querying multiple tables in a single statement, so I'm just kind of guessing at this point. Thanks in advance. |
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