Printable View
Ok, so I'm using the trim function on a field being retrieved from a MS SQL Database. The field appears to be empty, but the following does not work.It simply outputs "- -"... with a space between the dashes. Any ideas why?PHP Code:echo "<p class=warn>-" . trim($row[5]) . "-";
I am notgood at this. But you could try to put " " around the trim function like this:
PHP Code:echo "<p class=warn>-" . "trim($row[5])" . "-";
If that doesn't work, have you tested it without the trim function. Is it actually anything there?
Or you could try to do the trimming before the echo line...
What you posted should work. try:
echo $row[5]; //the value of $row[5]
or:
echo strlen($row[5]); //The length of the value.
I've tried all sorts of ways and used the strlen function. It is telling me that when there supposedly isn't anything there, the length of the string is 6 before and after the trim. Now I'm just really confused.
Does the data returned correlate with whats in the table?Quote:
Originally posted by ober5861
I've tried all sorts of ways and used the strlen function. It is telling me that when there supposedly isn't anything there, the length of the string is 6 before and after the trim. Now I'm just really confused.
I'd imagain you have made an error in retrieving the data from the database. Check your query and that the column you are expecting to see corresponds to the column in your sql query.
It does correlate. The field is a varchar of length 5.... and besides, it doesn't matter what it is returning... it's showing a single space both before and after the trim.
It only trims spaces at start and end of string. Here's of an example at work:
this shows you that the lenghts are different.PHP Code:<?
$a = array();
$a[0] = " a b c ";
$a[1] = " d e f ";
$a[2] = " g h i ";
echo strlen($a[2])."<br>";
echo strlen(trim($a[2]));
?>
If you get rid of the strlen() bit you'll see that they output the same anyway, but the source will look different.
Right... I understand that. But here is what is happening: I'm printing out the raw value with a dash in front and behind.It says that length is 6 using strlen();Code:"- -"
I then do a trim on what appears to be one space while strlen() says it's 6 and it outputs the SAME THING:Do you see my frustration?Code:"- -"
Oh I totally see your frustration.
try this and see what HTML the server outputs:
That'll tell you EXACTLY what row[5] is.PHP Code:echo "<p>".$row[5]."</p>";
No change, still gives me a space.
can you post the HTMl it gives? is it just:Quote:
Originally posted by ober5861
No change, still gives me a space.
"<p> </p>"?
and strlen($row[5]); is still 6? Then something VERY strange is going on.
Take a look at your HTML source, I think you may be trying to display a field with just spaces. If that is the case the browser will only display one space.
To see all six you'll need to convert the spaces to their HTML Entitiy equivilent: &nbsp;
Try this:
PHP Code:echo ('[value]' . htmlentities (trim ($row[5])) . ' [value]');
Acidic, that outputs "<p> </p>".
VisualAd, that outputs [value] [value].
More ideas?
if that is outputting:
Which you said, and also that:Code:<p> </p>
outputs 6. Which you also said. Then something damn wierd is going on. Are you sure $row is actually an array, try:PHP Code:echo strlen($row[5]);
That should tell you how many elements there are in the array. Also try:PHP Code:echo count($row);
Just to see what it outputs.PHP Code:echo $row;
It is apparent that the string does not contain any white space. Try the code below and it will tell you the ascii character code of each character in your string.
Maybe it will shed some light on your problem:
PHP Code:for ($count = 0; $count < str_len($row[5]); $count++)
echo (ord (sub_str ($row[5], $count, 1)) . '<br />');
I'm quite sure it is an array as I am able to output the rest of the elements ($row[0], $row[1], $row[2], etc.) just fine. It's like the browser won't display a blank field correctly or allow me to detect a blank field.
Ok... a light bulb just popped over my head and I have solved the problem... I've detected the content of $row[5] by saying:and that appears to work as I need it to.PHP Code:if($row[5] = " ")
Thanks for all your efforts.