Asked  7 Months ago    Answers:  5   Viewed   40 times

Is there a way to convert an integer to a string in PHP?

 Answers

19

You can use the strval() function to convert a number to a string.

From a maintenance perspective its obvious what you are trying to do rather than some of the other more esoteric answers. Of course, it depends on your context.

$var = 5;

// Inline variable parsing
echo "I'd like {$var} waffles"; // = I'd like 5 waffles

// String concatenation 
echo "I'd like ".$var." waffles"; // I'd like 5 waffles

// The two examples above have the same end value...
// ... And so do the two below

// Explicit cast 
$items = (string)$var; // $items === "5";

// Function call
$items = strval($var); // $items === "5";
Wednesday, March 31, 2021
 
alioygur
answered 7 Months ago
68

It depends on the driver used between php and mysql.

Check which one of them is used by checking pdo_mysql section of the output of

php -i

your output should be similar to

pdo_mysql

PDO Driver for MySQL => enabled
Client API version => mysqlnd 5.0.12-dev - 20150407 - $Id: b396954eeb2d1d9ed7902b8bae237b287f21ad9e $

The native driver return integers as integers, but the other return them as strings.

So the solution is to remove the old driver and install the native one.

or to use $casts into your model.

    protected $casts = [
    'status' => 'integer',
];
Wednesday, March 31, 2021
 
ramdemon
answered 7 Months ago
73

CStr(45) is all you need (the Convert String function)

Tuesday, July 27, 2021
 
turson
answered 3 Months ago
33

The following is probably not the neatest way, but it works:

1> lists:flatten(io_lib:format("~p", [35365])).
"35365"

EDIT: I've found that the following function comes in useful:

%% string_format/2
%% Like io:format except it returns the evaluated string rather than write
%% it to standard output.
%% Parameters:
%%   1. format string similar to that used by io:format.
%%   2. list of values to supply to format string.
%% Returns:
%%   Formatted string.
string_format(Pattern, Values) ->
    lists:flatten(io_lib:format(Pattern, Values)).

EDIT 2 (in response to comments): the above function came from a small program I wrote a while back to learn Erlang. I was looking for a string-formatting function and found the behaviour of io_lib:format/2 within erl counter-intuitive, for example:

1> io_lib:format("2 + 2 = ~p", [2+2]).
[50,32,43,32,50,32,61,32,"4"]

At the time, I was unaware of the 'auto-flattening' behaviour of output devices mentioned by @archaelus and so concluded that the above behaviour wasn't what I wanted.

This evening, I went back to this program and replaced calls to the string_format function above with io_lib:format. The only problems this caused were a few EUnit tests that failed because they were expecting a flattened string. These were easily fixed.

I agree with @gleber and @womble that using this function is overkill for converting an integer to a string. If that's all you need, use integer_to_list/1. KISS!

Sunday, August 8, 2021
 
Easen
answered 3 Months ago
15

This is either a bug in TypeScript or a concious design decision, but you can work around it using:

var myBool: bool = true;
var myString: string = String(myBool);
alert(myString);

In JavaScript booleans override the toString method, which is available on any Object (pretty much everything in JavaScript inherits from Object), so...

var myString: string = myBool.toString();

... should probably be valid.

There is also another work around for this, but I personally find it a bit nasty:

var myBool: bool = true;
var myString: string = <string><any> myBool;
alert(myString);
Friday, September 17, 2021
 
Pratap Vhatkar
answered 1 Month ago
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