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PHP Discussion Thread
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Jul 24th, 2009

string variable question

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hi, i selected everything from the table and the direct variable $cResult[3] would give me something like "street address with space as delimiter" ... however, when i tried to use it, it will give me only "street" instead of "street address with space as delimiter".

does anyone know how i can fix it so it will show the whole field?
thanks

PHP Syntax (Toggle Plain Text)
 
 
<?php
	require_once('auth.php');
	require_once('config.php');
 
	$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
	if(!$link) 
	{
		die('Failed to connect to server: ' . mysql_error());
	}
 
	//Select database
	$db = mysql_select_db(DB_DATABASE);
	if(!$db) 
	{
		die("Unable to select database");
	}
 
	$cid = $_SESSION['SESS_CUSTOMER_ID'];
 
	$qry="SELECT fname, lname, email, streetAddress,zip_id, passwd FROM customer WHERE customer_id='$cid'";
 
/////////////////// problem begin //////////////////////////////////	
        $cResult=mysql_fetch_array(mysql_query($qry));
 
	$fn= "$cResult[0]";
	$ln= "$cResult[1]";
	$email= "$cResult[2]";
	$street= "$cResult[3]";   //---> this will give me only part of the field (the first word basically)  
///////////////// problem end ///////////////////////////////

<?php
	require_once('auth.php');
	require_once('config.php');
	
	$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
	if(!$link) 
	{
		die('Failed to connect to server: ' . mysql_error());
	}
	
	//Select database
	$db = mysql_select_db(DB_DATABASE);
	if(!$db) 
	{
		die("Unable to select database");
	}
	
	$cid = $_SESSION['SESS_CUSTOMER_ID'];
	
	$qry="SELECT fname, lname, email, streetAddress,zip_id, passwd FROM customer WHERE customer_id='$cid'";

/////////////////// problem begin //////////////////////////////////	
        $cResult=mysql_fetch_array(mysql_query($qry));
	
	$fn= "$cResult[0]";
	$ln= "$cResult[1]";
	$email= "$cResult[2]";
	$street= "$cResult[3]";   //---> this will give me only part of the field (the first word basically)  
///////////////// problem end ///////////////////////////////

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k2k

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Posting Whiz

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Jul 24th, 2009

Re: string variable question

Try replacing the problem area with the following:

php Syntax (Toggle Plain Text)
$cResult=mysql_fetch_array(mysql_query($qry));
$fn=$cResult[0];
$ln=$cResult[1];
$email=$cResult[2];
$street=$cResult[3];
$cResult=mysql_fetch_array(mysql_query($qry));
$fn=$cResult[0];
$ln=$cResult[1];
$email=$cResult[2];
$street=$cResult[3];

cwarn23

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Jul 24th, 2009

Re: string variable question

Click to Expand / Collapse Quote originally posted by cwarn23 ...

Try replacing the problem area with the following:
php Syntax (Toggle Plain Text)
$cResult=mysql_fetch_array(mysql_query($qry));
$fn=$cResult[0];
$ln=$cResult[1];
$email=$cResult[2];
$street=$cResult[3];
$cResult=mysql_fetch_array(mysql_query($qry)); $fn=$cResult[0]; $ln=$cResult[1]; $email=$cResult[2]; $street=$cResult[3];

i will try again later but i think i already tried that and didn't work.

k2k

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Posting Whiz

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Jul 24th, 2009

-1

Re: string variable question

Just a thought, is your 4th field (street) long enough to take the whole street name. Check in phpmyadmin or whatever you use to admin your db to see if the data in 'street' is complete.

If this ain't a problem, have you got any strange characters following the first word?

ardav

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Jul 24th, 2009

Re: string variable question

Click to Expand / Collapse Quote originally posted by ardav ...

Just a thought, is your 4th field (street) long enough to take the whole street name. Check in phpmyadmin or whatever you use to admin your db to see if the data in 'street' is complete.

If this ain't a problem, have you got any strange characters following the first word?

nope, no strange characters. the field is long enough and the street i inserted "132 main street" was already stored in the table.

but it is weird that [code] echo "$cResult[3]" returns me only "132"

k2k

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Posting Whiz

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Jul 25th, 2009

Re: string variable question

Click to Expand / Collapse Quote originally posted by k2k ...

but it is weird that [code] echo "$cResult[3]" returns me only "132"

Perhaps it is an error in the part of the script that inserts into the table. Try using the mysql_real_escape_string() function to see if it escapes the spaces. And remember to put quotes around each string to be inserted. Below is an example:

php Syntax (Toggle Plain Text)
echo '`column`="'.mysql_real_escape_string('132 main street').'"';
mysql_query('INSERT INTO `table` SET `column`="132 main street"') or die(mysql_error());
echo '`column`="'.mysql_real_escape_string('132 main street').'"';
mysql_query('INSERT INTO `table` SET `column`="132 main street"') or die(mysql_error());

And check if the first line returns whats in the mysql query.

cwarn23

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Jul 25th, 2009

Re: string variable question

still don't know why, but it magically works now with $street= $cResult[3]

and echo "$cResult[3]" return "132 main street"

..... thanks everyone tried to help.

k2k

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Posting Whiz

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