I have a page with this global on it:
$sampleIssue = array('vol'=>25,'no'=>3 and 4);
on another page, I have this code which returns a 1 -???? What is wrong? I am sure it is something simple, but i am new enough to php not to know. I have searched white spaces in variables and strings, but still can't find the answer.
echo $sampleIssue['no'];
Change
$sampleIssue = array('vol'=>25,'no'=>3 and 4);
to
$sampleIssue = array('vol'=>25,'no'=>'3 and 4');
"3 and 4" is a string. You have to put it with in quotes. 25 will not show any problem because it is a number
$sampleIssue = array('vol'=>25,'no'=>3 and 4);
You need some quotes around 3 and 4, like
$sampleIssue = array('vol'=>25,'no'=>'3 and 4');
The expression 3 and 4 without quotes evaluates to Are both the number 3 and the number 4 "true"?, which itself is true. If you echo out a PHP boolean true, it displays as "1".
Related
Hello my dear coding friends.
I have a time, formatted like this 08:00:00. That time comes from my phpMyAdmin database where i have a "time" field and i get that field by using a query in my php code. The variable type of that mysqli variable containing the time is string, so i want to cut the minutes and seconds part off and turn the rest into an integer by adding (int). The code looks like this: Image of code
if (strpos ($meetings["dtStartZeit"], "0") == 0) {
$startTimeString = substr ($meetings["dtStartZeit"], 1, 1);
} else {
$startTimeString = substr ($meetings["dtStartZeit"], 0, 2);
}
$startTimeNumber = (int)$startTimeString;
Now comes the confusing part. If i have a string like this --> "8" and I want to turn it into an integer by using the above mentioned function, the result is 9 and not 8. The even more confusing part is that if I increase the value of that variable by 1, the result is 8.
Can someone explain me this please?
You don't need to use strpos or substr here. Use a single line type cast instead all of your code:
$startTimeNumber = (int) $meetings['dtStartZeit']; // "08:00:00" --> 8
To convert a string to an int you use intval()
because a string is an object of chars casting them wont ever work as expected which is what (int) is doing
Hi i need to save a 010 number in $number and if i do like this php will remove the starting 0
$number = 010
And echo of this will return 10 how can i make it not to remove the initial 0
BR
Martin
Use it as a String:
$number = '010';
Use str_pad() function.
echo str_pad('10',3,'0',STR_PAD_LEFT)
http://php.net/manual/en/function.str-pad.php
Do remember that numbers starting with 0 can also be treated as octal number notation by the PHP compiler, hence if you want to work with decimal numbers, simply use:
$num = '010';
This way the number is saved, can be stored in the database and manipulated like any other number. (Thx to the fact that PHP is very loosely typed language.)
Another method to use would be:
Save number as $num = 10;
Later while printing the value you can use sprintf, like:
sprintf("%03d", $i);
This will print your number in 3 digit format, hence 0 will be added automatically.
Another method:
<?php
$num = 10;
$zerofill = 3;
echo str_pad($num, $zerofill, "0", STR_PAD_LEFT);
/* Returns the wanted result of '010' */
?>
You can have a look at the various options available to you and make a decision. Each of the method given above will give you a correct output.
First I thought this was a stupid question, and i should do some search and it would be easy to solve. But I am afraid I just ain't getting anywhere!
The thing i need to do is simple. I have a U$ value and i want to divide it by 12. Thats it.
Well, the thing is that this value is outputed by a function, and echoes ok, look:
<?php
$preconormal = wpsc_the_product_price(); // it echoes like 99.90
$precoja = str_replace (".", "", $preconormal);
echo $precoja; //echo ok -> 9990
$quantas = '12';
$parcela = $precoja/$quantas; // ok, so divide 9990 by 12, right?
echo $parcela; //no!!!!! it echoes 0 :(
?>
I really hope you can help me!
You are trying to divide strings, if you used numbers say
$quantas = 12;
$precoja = 9990;
What happens?
It should fix the division, in which case, prior to the mathmematics, convert your vars to integs by
$quantas = intval($quantas);
$precoja = intval($precoja);
//your manipulation here..l
Remove the quotes...
$quantas = '12';
to
$quantas = 12;
$precoja = floatval($preconormal)*100;
$preconormal = $precoja / 12;
I'd change your 5th line by removing the single quotes and/or 6th line with $parcela = (int)$precoja / (int)$quantas; because as soon as you use the function str_replace then $precoja becomes a string. Also having the single quotes earlier on = '12' it is also a string and that division returns 0.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
PHP: show a number to 2 decimal places
How can I format an input number to be 0.00 if it has not any value? I tried (double) but it prints 0 only.
Here you go :)
echo number_format($var,2);
If you want it to print specific no. of decimal points, use number_format.
$float_var = number_format($var, 2);
$var = number_format($number, 2, '.', '');
This forces 2 points after the decimal, sets the decimal as a period. You can also forego the last two as it defaults to it;
Note: The third value is your decimal separator, the fourth value is the thousandths separator.
$var = number_format($number, 2);
For direct output:
printf('%0.2f',$var);
Output into variable:
$outVar = sprintf('%0.2f',$var);
This statemant casts $var type to float and prints with 2 decimal signs
maybe you should check it first if the value is not set
if(!isset($variableName))
{
// then set
$variableName = "0.00"; // => string
//or like this
$variableName = number_format(0,2); // => this result is also string
}
echo "value: ",$variableName;
result
0.00
if you are trying to format the value i sugest you to use meioMask pluging.
So you define your field as number and the pluging do the trick, even if you set "0" for the value
If I have a variable in PHP containing 0001 and I add 1 to it, the result is 2 instead of 0002.
How do I solve this problem?
$foo = sprintf('%04d', $foo + 1);
It would probably help you to understand the PHP data types and how they're affected when you do operations to variables of various types. You say you have "a variable in PHP say 0001", but what type is that variable? Probably a string, "0001", since an integer can't have that value (it's just 1). So when you do this:
echo ("0001" + 1);
...the + operator says, "Hm, that's a string and an integer. I don't know how to add a string and an int. But I DO know how to convert a string INTO an int, and then add two ints together, so let me do that," and then it converts "0001" to 1. Why? Because the PHP rules for converting a string to an integer say that any number of leading zeroes in the string are discarded. Which means that the string "0001" becomes 1.
Then the + says, "Hey, I know how to add 1 and 1. It's 2!" and the output of that statement is 2.
Another option is the str_pad() function.
$text = str_pad($text, 4, '0', STR_PAD_LEFT);
<?php
#how many chars will be in the string
$fill = 6;
#the number
$number = 56;
#with str_pad function the zeros will be added
echo str_pad($number, $fill, '0', STR_PAD_LEFT);
// The result: 000056
?>