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dadamssg

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Parse error: parse error in C:\wamp\www\PHPtutorials\showtoday.php on line 18

 

heres the whole code...im just trying to display all the info in tables

 

<?php
/* program: showevents.php*/

?>

<html>
<body>
<?php
   session_start();
   include("caneck.inc");
   
   $cxn = mysqli_connect($host,$user,$passwd,$dbname)
          or die ("Couldn't connect")
    

/*select todays events*/

$query = "SELECT * FROM test WHERE DATE(start) <= CURDATE() AND DATE(end) >= CURDATE() ORDER BY start DESC";

$result = mysqli_query($cxn,$query)
          or die ("Couldn't execute");
		  
while($row = mysqli_fetch_assoc($result))
   {    
    /*Format dates to display*/

     $startt = DATE_FORMAT($row['start'], '%b %e %Y %l %i %p');
 $endt = DATE_FORMAT($row['end'], '%b %e %Y %l %i %p');

echo " <table border='3'>";
echo " <tr> 
echo "<td>{$row['eventid']}</td>";
echo "<td>{$row['created']}</td>";
echo "<td>{$row['title']}</td>";
echo "<td>{$row['event']}</td>";
echo "<td>{$row['description']}</td>";
echo "<td>{$row['location']}</td>";
echo "<td>\$$startt</td>";
echo "<td>\$$endt</td>";
echo "</tr></table>";
   }
?>
</body></html>


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That's a php parse error. A lot of php parse errors are due to a problem in the code prior to where the error is reported. The actual error cannot be detected because the parser does not know what you intended and it is only when it encounters something that is out of place in the current context that it generates the error message.

 

You are missing a semi-colon ; on the line of code immediately prior to where the error is being reported.

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yeah i fixed that..nothin. I don't get it all... i am copying and pasting inside the quotes of the $query and putting them inside PhpMyAdmin and it works. this is driving me crazy

 

<?php
/* program: showevents.php*/

?>

<html>
<body>
<?php
   session_start();
   include("caneck.inc");
   
   $cxn = mysqli_connect($host,$user,$passwd,$dbname)
          or die ("Couldn't connect");
    
/*select todays events*/

$query = "SELECT * FROM test WHERE DATE(start) <= CURDATE() AND DATE(end) >= CURDATE() ORDER BY start DESC";

$result = mysqli_query($cxn,$query)
          or die ("Couldn't execute");
		  
while($row = mysqli_fetch_assoc($result))
   {    
    /*Format dates to display*/

     $startt = DATE_FORMAT($row['start'], '%b %e %Y %l %i %p');
 $endt = DATE_FORMAT($row['end'], '%b %e %Y %l %i %p');

echo " <table border='3'>";
echo " <tr>"; 
echo "<td>{$row['eventid']}</td>";
echo "<td>{$row['created']}</td>";
echo "<td>{$row['title']}</td>";
echo "<td>{$row['event']}</td>";
echo "<td>{$row['description']}</td>";
echo "<td>{$row['location']}</td>";
echo "<td>\$$startt</td>";
echo "<td>\$$endt</td>";
echo "</tr></table>";
   }
?>
</body></html>





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ok so i guess php didn't like the names of my variables so i changed them and it almost worked, now im getting a datetime error when i try to format the dates...the dates are in datetime in the database i dunno why im getting this error now...ive never formatted dates that come from datetime though

 

the error:

Warning: date_format() expects parameter 1 to be DateTime, string given in showtoday.php on line 25

 

Warning: date_format() expects parameter 1 to be DateTime, string given in showtoday.php on line 26

 

im also getting a session error..i dunno why thats comin up either:

Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at showtoday.php:8) in showtoday.php on line 9

 

 

 

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