[SOLVED] str_replace in a loop

[dsaba]

here's a sample code of what i'm doing in simple terms:

<?php
//queryrow = mysql_fetch_array yields 10 rows in db

$string = 'Hello my name is Bob';

while ($queryrow = mysql_fetch_array($query)) 	{
$lookingfor = $queryrow['text'];
$replacedtext = $queryrow['othertext'];
$newstring = str_replace($lookingfor, $replacedtext, $string);
								}
echo $newstring;

//newstring echoes to the last replaced string, all the previous replacements are still there
?>

 

the problem is newstring is only replaced with the last str_replace, and the string does not have all the 10 things i replaced in it

 

how I do this with this loop?

-thanks

 

 

if you dont understand my code and example

image this:

 

you have an array of the numbers, 1, 2, 3, 4, 5, 6

you have a string of "one two three four five six"

 

in the database its setup like this:

numerical    spelled_out

1                one

2                two

3                  three

4                  four

5                  five

6                  six

 

so you make a while statement for each row of the database (thats a loop that loops 6 times!)

 

you're text string is "I'm 4 years old and I ran up 6 steps!"

 

you want the text string to become:

"I'm four years old and i ran up six steps"

 

you want to do this utilizing your database where you have definitions of numerical vs. spelled-out

 

how can it be done?

[paul2463]

here's a sample code of what i'm doing in simple terms:

<?php
//queryrow = mysql_fetch_array yields 10 rows in db

$string = 'Hello my name is Bob';

while ($queryrow = mysql_fetch_array($query)) 	{
$lookingfor = $queryrow['text'];
$replacedtext = $queryrow['othertext'];
$newstring = str_replace($lookingfor, $replacedtext, $string);
								}
echo $newstring;

//newstring echoes to the last replaced string, all the previous replacements are still there
?>

 

everything is fine, it is doing what you ask it to, here is what I mean

 

$string = "I'm 4 years old and I ran up 6 steps!"

 

and we have what you had

in the database its setup like this:

numerical    spelled_out

1                one

2                two

3                three

4                four

5                five

6                six

 

first time through the loop, nothing, second, nothing, third nothing fourth we get

$string = "I'm 4 years old and I ran up 6 steps!"

$newstring = str_replace($lookingfor, $replacedtext, $string); // becomes "I'm four years old and I ran up 6 steps!"

 

you make no changes to $string ******important******

 

next time round nothing then the sixth we get

$newstring = str_replace($lookingfor, $replacedtext, $string); // becomes "I'm 4 years old and I ran up six steps!"

 

*******BECAUSE YOU MADE NO CHANGES TO $string IT WORKS ON THE ORIGINAL STRING AGAIN*******

 

try this

<?php
//queryrow = mysql_fetch_array yields 10 rows in db

$string = 'Hello my name is Bob';

while ($queryrow = mysql_fetch_array($query)) 	
{
$lookingfor = $queryrow['text'];
$replacedtext = $queryrow['othertext'];
$string = str_replace($lookingfor, $replacedtext, $string);
}
echo $string;

//make the changes to the original string and keep changing that
?>

[dsaba]

wow thanks for going to explain it very well

i knew what was going on that was wrong, how i kept referencing the original string

 

i just couldn't put my finger on it on how to fix it, you did it quite simply

 

p.s. how did you highlight php code without putting it in the code brackets??