jobs1109

Hi everyone i got it solved. I just turned off magic quotes in php.ini Thanks

Hi everyone, I am having a problem with using quotation marks / apostrophe with php and mysql. If I enter the following lines in mysql. "user's resume" when I do a query from mysql i get back " User\'s resume" it's adding a slash before the apostrophe. How can I fix this ? Thanks.

How do I modify it so that it can be displayed mutiple times and still show the correct login/logout option ?

Hi, I am having a bit of problem with my login/logout script. When user is logged in I want the script to show logout and if they are not login I want the script to show login. The problem is even when the user is logged in it says "you must be logged in Click here to login " here is the script Please help <?php session_start(); $_SESSION['username'] = $_POST['username']; if ($_SESSION['username']) echo "Welcome, ".$_SESSION['username']."!<br><a href='logout.php'>Logout</a>" ; else die("you must be logged in <a href='Login.php'>Click here to login</a>"); ?> what did I do wrong in this script ? Thanks

Thanks for your help I will try it.

Hi yes employer are assigned to a company and jobs are assigned to a company. I am not sure how to implement the relationships.

Hi I got three tables (employers , company , and Jobs) Employer table holds info about employer. Company hold info about the company and jobs table holds info about jobs. I was just wandering what would be a good way to link these tables in the database ? Here is the coding for each table. Employer Table CREATE TABLE IF NOT EXISTS `employers` ( `id` int(11) NOT NULL, `username` varchar(50) NOT NULL, `password` varchar(50) NOT NULL, `email` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Jobs Table CREATE TABLE IF NOT EXISTS `Jobs` ( `id` int(11) NOT NULL, `JobTitle` varchar(200) default NULL, `Company` varchar(200) default NULL, `Salary` varchar(30) default NULL, `Description` varchar(2000) default NULL, `CompanyURL` varchar(200) default NULL, `PhoneNumber` varchar(30) default NULL, `Requirements` varchar(2000) default NULL, `JobCategory` varchar(100) default NULL, `JobType` varchar(100) default NULL, `Apply_To` varchar(1000) NOT NULL, `Email` varchar(200) NOT NULL, `modified_at` datetime NOT NULL, `PostedOnDate` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP, `Address` varchar(250) NOT NULL, `State` varchar(200) NOT NULL, `City` varchar(200) NOT NULL, `Country` varchar(100) NOT NULL, `Zipcode` varchar(100) NOT NULL, `JobID` varchar(100) NOT NULL, `WorkExperience` varchar(2000) NOT NULL, `EducationRequirement` varchar(2000) NOT NULL, `WebsitePostedFrom` varchar(200) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; Company Table CREATE TABLE IF NOT EXISTS `Company` ( `id` int(11) unsigned NOT NULL auto_increment, `CompanyName` varchar(100) NOT NULL, `Address` varchar(100) NOT NULL, `Logo` varchar(100) NOT NULL, `PhoneNumber` varchar(25) NOT NULL, `ContactPerson` varchar(25) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;

Hi thank you very much it's working now.

Here is the structure for the table -- -- Table structure for table `Company` -- CREATE TABLE IF NOT EXISTS `Company` ( `id` varchar(25) NOT NULL, `CompanyName` varchar(100) NOT NULL, `Address` varchar(100) NOT NULL, `Logo` varchar(100) NOT NULL, `PhoneNumber` varchar(25) NOT NULL, `ContactPerson` varchar(25) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8;

Hi I am getting this error message and don't know why. please help this is the message " Error: Duplicate entry '' for key 1 " Here is the code <?php include'../DB-connection.php'; $sql="INSERT INTO Company(CompanyName,Address,Logo,PhoneNumber,ContactPerson) VALUES('".mysql_real_escape_string($_POST[CompanyName])."', '".mysql_real_escape_string($_POST[Address])."', '".mysql_real_escape_string($_POST[Logo])."', '".mysql_real_escape_string($_POST[PhoneNumber])."', '".mysql_real_escape_string($_POST[ContactPerson])."')"; if(!mysql_query($sql,$con)) { die(' Error: '. mysql_error()); } echo " <center>your compnay info added &nbsp &nbsp &nbsp <a href='User-Login.php'>Login to Post a job</a> </center>" ; mysql_close($con) ?>

Hi, I am thinking about starting a coupon website. I would like to use php and mysql. I was wander if there is any tutorial on how to go about coding a coupon website ? Thanks

both users TABLE and Jobs TABLE =InnoDB

Hi when I remove the brackets I get the followin message. Here is the code ALTER TABLE Jobs ADD FOREIGN KEY jointable REFERENCES users (id); here is the error message Error SQL query: ALTERTABLE Jobs ADDFOREIGNKEY jointable REFERENCES users( id ) MySQL said: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'REFERENCES users (id)' at line 1

Hi everyone, I am trying to add a foreign key (jointable) to an existing table called (Jobs) and I would like it to reference the id in users table. Also I like to have it auto update when records added and deleted. ENGINE=InnoDB When I run the following code I get an error message. ALTER TABLE Jobs ADD FOREIGN KEY (jointable) REFERENCES users (id); here is the error message Error SQL query: ALTERTABLE Jobs ADDFOREIGNKEY ( jointable )REFERENCES users( id ) MySQL said: #1072 - Key column 'jointable' doesn't exist in table What am I doing wrong ? please help

Hi , Here is what I got so far , a form where users can login and post thei jobs now I would like to have them edit their own postings. http://www.alwayshawaii.com/Sawh-Network/Hilo-Jobs/User-Login/User-Login.php

How could I get users to edit, delete, and update only their own postings ?

Hi Thanks but I am looking for a tutorials that goes a bit more on how to store data using sessions in a database and then be able to edit, update, delete using sessions.

Hi, I am looking for a session tutorial where users can save data to a mysql database and then edit / update/delete only their postings. If you know of a good tutorial please let me know. Thanks D.S

Hi thanks for helping but I am getting an error message. Here is the code i have now <center> <?php do{ <?php echo '<img border="0" src="http://www.hiringinhilo.com/Hilo-Jobs/Hilo-Logo/job-search-results-top.gif">'; echo '<table width="500">'; echo'<tr><th>Job Title</th><th>Date Posted</th><th>State</th><th>Job Type</th></tr>';while ($rsjobs=mysql_fetch_assoc($jobs_query)){echo '<tr><td>'; echo '<td><a href="http://www.hiringinhilo.com/Hilo-Jobs/Job-Search-Form/Getting-Data-From-Database/Job-Details/Job-Details.php?id=<'.$rsjobs['id'].' target="new">'.$rsjobs['JobTitle'].'</a></td>';$date = date("m/d/Y",strtotime($rsjobs['PostedOnDate'])); echo '<td>'.$date.'</td>'; echo '<td>'.$rsjobs['State'].'</td>'; echo '<td>'.$rsjobs['JobType'].'</td>';echo '</tr>';}echo '</table>'; } ?> </center> and here is the error message Parse error: syntax error, unexpected '<' in /home/content/h/i/l/hilo103/html/Hilo-Jobs/Job-Search-Form/Getting-Data-From-Database/Getting-Data-Database4.php on line 51

Hi, I am trying to get this Mysql results in a table Here is the output <table width="500"> <tr><td><center><img border="0" src="http://www.hiringinhilo.com/Hilo-Jobs/Hilo-Logo/job-search-results-top.gif"><center><br><br></td></tr> <tr><td> <?php do {?> <p><a href="http://www.hiringinhilo.com/Hilo-Jobs/Job-Search-Form/Getting-Data-From-Database/Job-Details/Job-Details.php?id=<?php echo $rsjobs['id'];?>" target="new"><?php echo $rsjobs['JobTitle']; ?></a> <?php echo date("m/d/Y",strtotime($rsjobs['PostedOnDate'])); ?> <?php echo $rsjobs['State'];?> <?php echo $rsjobs['JobType'];?> <?php } while ($rsjobs=mysql_fetch_assoc($jobs_query)) ?> </td></tr> </table> </center> How do I get this to print out in a table like Job title Date Posted State Job type Thanks

Hi Thanks this code work. Thank you very much.

<?php echo $rsjobs['PostedOnDate'];?> time show usp as 2011-08-26 04:12:57 I would like to get it to show up as 08/26/2011

Hi here is how it's being retreived from database " <?php echo $rsjobs['PostedOnDate'];?> $rsjobs is the id corresponding to that row of data. How do I format this to "08/26/2011 " ? "

Thanks that works. Now when I get the results from the mysql query and print it out it show as " 2011-08-26 04:07:23 " How do I get it to show as 08/26/2011 ?

Hi Everyone, I am trying to create a variable with the current date and then insert it into a mysql database. Here is what I have so far. $DatePostedOn= date("F j, Y, g:i a"); when I insert the variable into the table it shows " 0000-00-00 00:00:00 " What am I doing wrong ? Please help