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gristoi

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Posts posted by gristoi

    sql function error

    in PHP Coding Help

    Posted

    Change you query and remove the brackets around the pid and you need to alias the count:

    "SELECT COUNT('post_id') AS 'postIdCount' FROM 'posts' WHERE 'post_id' = $pid"

     

    the extra brackets in the query are terminating your script further on

    sql function error

    in PHP Coding Help

    Posted

    Hi, like I stated previously your have some syntax errors within your mysql query. firstly, if you are joining tables together in a query then you need to use the concatenation operator '.' to prepend the tablename to the field name. In your query you were using a comma to do this.

     

    Secondly, you do not need to surround the fieldnames in the query with single quote marks, so posts.postid instead of 'posts'.'postid' is fine. You only need to use '' to surround the alias name of you fieldname - 

    posts.postid AS 'ID'

    . There wil however be occasions when you do need to ensure there is not a conflict with reserved MYSQL words and for this you can surround the name with ``.

     

    I did miss some of the commas on your query before but it should look like this:

    SELECT
			posts.post_id AS 'id',
			posts.post_title AS 'title',
			LEFT(posts.post_body,'512') AS 'preview',
			posts.post_user AS 'user',
			DATE_FORMAT(posts.post_date,'%d/%m/%Y %H:%i:%s') AS 'date',
			comments.total_comments,
			DATE_FORMAT(comments.last_comment,'%d/%m/%Y %H:%i:%s') AS 'last_comment'
		FROM posts
		LEFT JOIN (
			SELECT
				post_id,
				COUNT('comment_id') AS 'total_comments',
				MAX('comment_date') AS 'last_comment'
			FROM comments
			GROUP BY post_id
		) AS comments
		ON posts.post_id = comments.post_id
		ORDER BY posts.post_date DESC;

     

    A bit of advice with working with your queries: to check whether the query is syntactically correct ( and works ) I always directly run my query in my mysql workbench / phpmyadmin panel or sql command line on my dev environment to ensure that it is working properly. If it is then that allows me to look elswhere for what could be causing the issue. Tools such as mysql workbench allow you to write and test your queries and will highlight if there is a syntax error in your query, very useful.

     

    The looping error you are getting is more than likely due to the syntax error in your query not returning a valid resource.

     

    Hope this helps

    php-soap

    in PHP Coding Help

    Posted 

    print($client->RetrieveArticles("Article"));

     

    Is incorrect, RetrieveArticles is returning an array. Try:

    $result = $client->RetrieveArticles("Article");
print_r($result);

     

    And see what it gives you

    Parse error: syntax error, unexpected '{' Error Help

    in PHP Coding Help

    Posted

    I dont understand what you are on about.

     

    the issue you had was with 

    while($row = mysql_fetch_assoc($out);  <--- you had a semi colon and were missing a ')'

     

    So i fixed that and reported your code.

     

    So please read through you code before saying you already know. If you 'already knew' then you wouldnt have posted your issue on the forum

    sql function error

    in PHP Coding Help

    Posted 

    SELECT
			posts.post_id AS 'id',
			posts.post_title AS 'title',
			LEFT(posts.post_body,'512') AS 'preview',
			posts.post_user AS 'user',
			DATE_FORMAT(posts.post_date,'%d/%m/%Y %H:%i:%s') AS 'date',
			comments.total_comments,
			DATE_FORMAT(comments.last_comment,'%d/%m/%Y %H:%i:%s') AS 'last_comment'
		FROM posts
		LEFT JOIN (
			SELECT
				post_id,
				COUNT('comment_id') AS 'total_comments',
				MAX('comment_date') AS 'last_comment'
			FROM comments
			GROUP BY post_id
		) AS comments
		ON posts.post_id = comments.post_id
		ORDER BY 'posts','post_date' DESC;

    Parse error: syntax error, unexpected '{' Error Help

    in PHP Coding Help

    Posted

    try:

    <?php
mysql_connect ("-","-","-") or die ('Error');
mysql_select_db ("-");

$out = mysql_query("SELECT * FROM guestbook ORDER BY id DESC");

while($row = mysql_fetch_assoc($out))
{
$name = $row['name'];
$email = $row['email'];
$txt = $row['comment'];

        $msg = "Are you sure you want to delete";

/* @var $_REQUEST <type> */
if (isset($_REQUEST ["action"]) && $_REQUEST["action"] == "del") 
    {
                    $id = intval($_REQUEST['id']);
                    mysql_query("DELETE FROM guestbook WHERE id=$id;");
                   echo "<action=index.php>";
                }

echo "<font face='verdana' size='1'>";
echo "<table border='0'>
		<tr><td>Name: ".$name."</td></tr>"."
		<tr><td>Email: ".$email."</td></tr>
		<tr><td colspan='2'>Comment:</td></tr>
		<tr><td colspan='2' width='500'><b>".$txt."</b></td></tr>
                     <tr><td><a onclick=\"return confirm('.$msg.');\" href='index.php?action=del&id=".$row['id']."'><span class='red'>["."Delete"."]</span></a>
                        </td></tr>
	  </table><br />";


echo "<hr size='1' width='500' align='left'></font>";
}
?>

    Code to email form data

    in PHP Coding Help

    Posted

    Hi, sorry, I never had time to look over all of the code before. You have the email function in the wrong part of your script, all of the email code you have written needs to go directly after your update query and before the 

    $URL =[\code] line. The errors you are getting is because it is sitting on a part of the script that hasn't been posted yet.

    Code to email form data

    in PHP Coding Help

    Posted

    It might not be throwing an error. It might be a logic problem where your code will never hit that loop. One way to test is to manually kill the script with the 

    die('I am here');

     

    place that firstly before you call the mail part. If you see 'I am here' then if loop has been entered. Then place it after the mail function and see if it is called again. Im sure you get the picture

    Code to email form data

    in PHP Coding Help

    Posted

    you are trying to mail twice, once before you declare your headers. from line 145 change:

    	mail($sEmailAddress, $sSubject, $sMessage, $sHeaders);

	$sMessage = "<p>Someone has edited the information below.</p><br> Location : {$_REQUEST['location']}<br>First Name : {$_REQUEST['fname']}<br>Last Name : {$_REQUEST['lname']}";
	$sHeaders  = "MIME-Version: 1.0\n"; 
	$sHeaders .= "Content-type: text/html; charset=iso-8859-1\n"; 
	$subject  = "Please Review";
	$sHeaders .= "To: Supervisor <$sEmailAddress>\n"; 
	$sHeaders .= "From: The Database <sender email>\n"; 

	if (mail($sEmailAddress, $sSubject, $sMessage, $sHeaders)) {

	echo "<center>An email has been sent to the supervisor for review"; 

	} else { 

	echo "This system is not working properly. Please contact a tech."; 

	}

     

    To:

    	$sMessage = "<p>Someone has edited the information below.</p><br> Location : {$_REQUEST['location']}<br>First Name : {$_REQUEST['fname']}<br>Last Name : {$_REQUEST['lname']}";
	$sHeaders  = "MIME-Version: 1.0\n"; 
	$sHeaders .= "Content-type: text/html; charset=iso-8859-1\n"; 
	$subject  = "Please Review";
	$sHeaders .= "To: Supervisor <$sEmailAddress>\n"; 
	$sHeaders .= "From: The Database <sender email>\n"; 
        $sendMail = mail($sEmailAddress, $sSubject, $sMessage, $sHeaders);
	if (sendMail) {

	echo "<center>An email has been sent to the supervisor for review"; 

	} else { 

	echo "This system is not working properly. Please contact a tech."; 

	}

     

    also echo out 

    $sMessage

    and  

    $sHeaders

    to make sure you havent got anything escaping the string. If you are still getting no confirmation message  then this part of the loop is not being entered on the page and you will need to do some fault finding to find where the script is stopping.

    Newbie In Need of Assistance

    in PHP Coding Help

    Posted

     

    try:

    <table border="1" cellspacing="1" align="center" bordercolor="#FFFFFF">
<tr>
<td>Course Name</td>
<td>Edit</td>
<td>Submit</td>
<td>Delete</td>
</tr>
<?php
$query = "SELECT * FROM Courses ";
$result = mysql_query ($query);
if(mysql_num_rows($result)>0){
                while($row = mysql_fetch_array($result)) {
                echo'<form action="editcourses.php" method="post">';
                echo('<tr>' .
                '<td>' . $row['CourseName'] . '</td>' .
                '<td><input type="hidden" name="id" value="' . $row['ID'] . '"/><textarea name="coursenametext" rows="2"></textarea>' . '</td>' .
                '<td>' . '<input type="Submit" name="Submit" value="Submit">' . '</td>' .
                '<td>' . '<input type="Submit" name="Delete" value="Delete">' . '</td>' .
                '</tr>'
                '</form>';
);
        }
}
if(isset($_POST['Submit']))
{
$query1 = "UPDATE courses SET courseName = '".$_POST['coursenametext']."' WHERE ID = '".$_POST['id']."'";
$result1 = mysql_query ($query1);
}
if(isset($_POST['Delete']))
{
$query1 = "Delete FROM  courses WHERE ID = '".$_POST['id']."'";
$result1 = mysql_query ($query1);
}
?>
</table>   

     

    and as for adding a new one, you just need to create a new form outside of the loop and name the button create etc......

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