AdRock Posted July 19, 2006 Share Posted July 19, 2006 I am having problems displaying data from a database onto a page. i haven't used php for months so am getting back into it slowly.I am sure I have been able to add a record becuase everything seems fine but the page is displaying nothing when i try to view the content. Is there a way i can see what records exist in the database?Here is how i insert a new record[code]<form action="index.php?page=insertnews" method="post"> <p class="style1">Please enter a title for the news item. <input style="width:400px;" type="text" size="50" name="title"></p> <p class="style1">Please enter the content for the news item. <textarea style="width:400px;" name="content" cols="30" rows="10"> </textarea></p> <input class="submit-button" style="margin-left:0" type="Submit" value="Submit"><input class="submit-button" style="margin-left:2px" type="reset" value="Reset"></form>[/code]this is insertnews.php[code]<?php include_once("../includes/connection.inc"); $title= $_POST['title']; $content= $_POST['content']; mysql_connect($host,$user,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO news VALUES ('','$title','$content')"; mysql_query($query); mysql_close();?>[/code]And this is the page that the content is to be displayed on. I use a switch ($_GET['page']) to open pages [code]<h3><?php include_once("includes/connection.inc"); mysql_connect($host,$user,$password); @mysql_select_db($database) or die( "Unable to select database"); $result = @mysql_query("SELECT title FROM news"); if (!$result) { echo("<p>Error performing query: " . mysql_error() . "</p>"); exit(); } while ( $row = mysql_fetch_array($result) ) { echo("<p>" . nl2br($row["title"]) . "</p>"); }?></h3><p class="style3"><?php include_once("includes/connection.inc"); $result = @mysql_query("SELECT content FROM news"); if (!$result) { echo("<p>Error performing query: " . mysql_error() . "</p>"); exit(); } while ( $row = mysql_fetch_array($result) ) { echo("<p>" . nl2br($row["content"]) . "</p>"); } mysql_close();?></p>[/code] Quote Link to comment https://forums.phpfreaks.com/topic/15043-problems-displaying-data-from-database/ Share on other sites More sharing options...
Ninjakreborn Posted July 19, 2006 Share Posted July 19, 2006 is it blank, if it is then is error viewing off in php.ini if so it's syntax error, is it returning a specific error or what. Quote Link to comment https://forums.phpfreaks.com/topic/15043-problems-displaying-data-from-database/#findComment-60485 Share on other sites More sharing options...
SammyP Posted July 19, 2006 Share Posted July 19, 2006 The Insert query looks OK but I prefer the Set syntax to specifically set the fields. So long as your table has three fields should be fine though. I take it after you see the insertnews page you either redirect to the viewing page or you are doing it manually at the moment.Update table Set field1='sam', field2=16Other than that I don't know why you are dereferencing the mysql_query function with the @ symbol. I don't know if there is a reason for it but I never do it. Try it without and see if that works. Do you use phpMyAdmin or something similar to help set up and view your database?Sam. Quote Link to comment https://forums.phpfreaks.com/topic/15043-problems-displaying-data-from-database/#findComment-60535 Share on other sites More sharing options...
AdRock Posted July 20, 2006 Author Share Posted July 20, 2006 I've installed php etc on my local machine and set up my databaseI changed the form action from [code]<form action="index.php?page=insert" method="post">[/code] to [code]<form action="insert.php" method="post">[/code] and now it works i can insert records and I can display them.Is there no way I can use the switch statement so i can use the original form action?Here is the switch statement on the index.php page[code]<?php switch ($_GET['page']) { case "contact": include('contact.php'); break; case "news": include('news.php'); break; case "news1": include('news1.php'); break; default: include('home.php'); } ?>[/code]If there is no way around it i'll have to stick with it Quote Link to comment https://forums.phpfreaks.com/topic/15043-problems-displaying-data-from-database/#findComment-60957 Share on other sites More sharing options...
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