update record

[angel55]

I keep getting this error.plz help

 

Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in C:\xampp\htdocs\Estate\update_agent.php on line 19

 

I have highlighted line 19 below

<?php

$db1 = new agent();

$db1->openDB();

$sql="select Agent_ID from agents ";

$result=$db1->getResult($sql);

if (!$_POST)

{

?>

<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">

Select Agent ID to update: <select name="pid">

<?php

while($row = mysql_fetch_assoc($result))

echo "<option value='{$row['Agent_ID']}'>{ $row['Agent_ID']} </option>";?>

</select>

<input type="submit" value="Go" />

</form>

<?php

} // end if

 

elseif (!$_POST['Agent_name'] && !$_POST['Address '] && !$_POST['Postcode'] )

{

$Agent_ID = $_POST['Agent_ID'];

$sql="SELECT Agent_name,Address,Postcode, from agents where Agent_ID ={$Agent_ID}";

$result=$db1->getResult($sql);

$row = mysql_fetch_assoc($result);

if($row)

{

 

$Agent_ID = $row['Agent_ID'];

$Agent_name = $row['Agent_name'];

$Address= $row['Address'];

$Postcode = $row['Postcode'];

 

 

} // end of inner if

else

die("SQL Error: " . mysql_error());

$sql2 = "";

$result2=$db1->getResult($sql2);

if(!$result2)

{

die("SQL Error: " . mysql_error());

}

?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

Agent_ID:<input type="text" name="pid" value="<?php echo $Agent_ID; ?>" /><br />

Agent Name:<input type="text" name="name" value="<?php echo $Agent_name; ?>" /><br />

Address:<input type="text" name="name" value="<?php echo $Address; ?>" /><br />

Postcode:<input type="text" name="name" value="<?php echo $Postcode; ?>" /><br />

 

 

 

 

<?php

} // end of else if

else //user has submitted form

{

$Agent_ID = $_POST['Agent_ID'];

$Agent Name = $_POST['Agent Name'];

$Address = $_POST['Address'];

$Postcode= $_POST['Postcode'];

$numofrows = $db1->update($Agent_ID,$Agent Name,$Address,$Postcode);

echo "Success. Number of rows affected: <strong>{$numofrows}<strong>";

$db1->closeDB();

}

?>

[cunoodle2]

I'm guessing the issue is with the brackets "{ }" that you are using.  Try this..

 

<?php
while($row = mysql_fetch_assoc($result))
{
echo "<option value='$row['Agent_ID']'> $row['Agent_ID'] </option>";
}
?>

 

If that does work then try this...

<?php
while($row = mysql_fetch_assoc($result))
{
echo "<option value='".$row['Agent_ID']."'> ".$row['Agent_ID']." </option>";
}
?>

 

The first method is preferred and will run/render faster.  Also please enclose all of your code with the "code" tags.

[angel55]

that error is solved. ive got this error message now

Parse error: parse error in C:\xampp\htdocs\Estate\update_agent.php on line 68

 

i ave highlighted below

 

<?php

$db1 = new agent();

$db1->openDB();

$sql="select Agent_ID from agents ";

$result=$db1->getResult($sql);

if (!$_POST)

{

?>

<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">

Select Agent ID to update: <select name="pid">

<?php

while($row = mysql_fetch_assoc($result))

{

echo "<option value='".$row['Agent_ID']."'> ".$row['Agent_ID']." </option>";

}

?>

</select>

<input type="submit" value="Go" />

</form>

<?php

} // end if

 

elseif (!$_POST['Agent_name'] && !$_POST['Address '] && !$_POST['Postcode'] )

{

$Agent_ID = $_POST['Agent_ID'];

$sql="SELECT Agent_name,Address,Postcode, from agents where Agent_ID ={$Agent_ID}";

$result=$db1->getResult($sql);

$row = mysql_fetch_assoc($result);

if($row)

{

 

$Agent_ID = $row['Agent_ID'];

$Agent_name = $row['Agent_name'];

$Address= $row['Address'];

$Postcode = $row['Postcode'];

 

 

} // end of inner if

else

die("SQL Error: " . mysql_error());

$sql2 = "";

$result2=$db1->getResult($sql2);

if(!$result2)

{

die("SQL Error: " . mysql_error());

}

?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

Agent_ID:<input type="text" name="pid" value="<?php echo $Agent_ID; ?>" /><br />

Agent Name:<input type="text" name="name" value="<?php echo $Agent_name; ?>" /><br />

Address:<input type="text" name="name" value="<?php echo $Address; ?>" /><br />

Postcode:<input type="text" name="name" value="<?php echo $Postcode; ?>" /><br />

 

 

 

 

<?php

} // end of else if

else //user has submitted form

{

$Agent_ID = $_POST['Agent_ID'];

$Agent Name = $_POST['Agent Name'];$Address = $_POST['Address'];

$Postcode= $_POST['Postcode'];

$numofrows = $db1->update($Agent_ID,$Agent Name,$Address,$Postcode);

echo "Success. Number of rows affected: <strong>{$numofrows}<strong>";

$db1->closeDB();

}

?>

 

[cunoodle2]

Your code was..

$Agent Name = $_POST['Agent Name'];

 

change it to...

$Agent_Name = $_POST['Agent Name'];

 

OR..

$AgentName = $_POST['Agent Name'];

 

Php variables can't have spaces and SERIOUSLY next time use the "code" brackets or I (and probably along with other people here) will simply ignore your post.

[angel55]

thnx ive done tat now i got this error message.

 

Iam really new to php.do you know what the following error message means?

 

SQL Retrieve Error: 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 'from agents where Agent_ID =' at line 1

 

 

my sql table for the agent is below:

CREATE TABLE IF NOT EXISTS `agents` (

  `Agent_ID` int(4) NOT NULL AUTO_INCREMENT,

  `Agent_Name` varchar(20) DEFAULT NULL,

  `Address` varchar(40) DEFAULT NULL,

  `Postcode` varchar(7) DEFAULT NULL,

  PRIMARY KEY (`Agent_ID`)

) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;

 

 

[anupamsaha]

Your form element is named as "name"

 

Agent Name:<input type="text" name="name" value="<?php echo $Agent_name; ?>" />

 

So, change as follows:

 

$Agent_Name = $_POST['name'];

 

And, PHP and any programming language or script does not allow SPACE in a variable name.

 

[angel55]

is still not working.

 

giving me the same error

[anupamsaha]

is still not working.

 

giving me the same error

 

Why there are 3 identical name attributes to 3 different form input elements?

 

Agent Name:<input type="text" name="name" value="<?php echo $Agent_name; ?>" /><br />
Address:<input type="text" name="name" value="<?php echo $Address; ?>" /><br />
Postcode:<input type="text" name="name" value="<?php echo $Postcode; ?>" /><br />

 

Change them and change PHP variables accordingly.

 

[angel55]

i have changed them but still no luck