Using php to Auto generate C++ code

Question

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I have kind of a big application with a lot of values that are used as options. The control for the application is a website, application itself is C++ and all the options are stored inside MySQL so that both C++ and php can access it. So I am using php to read bunch of values from a text file, and auto generate SQL, C++ and php code, since the options are used everywhere and are so massive that it would take too long to implement them, and would be too much of a hassle to add an extra options in the future.

This is what one line of my text file looks like

`input_video_standard`|:|INT|:|NOT NULL DEFAULT '0'|:|profile->input.videoStandard|:|Video Standard|:|SD 576i@50Hz (B,G/PAL)|:|SD [email protected] (NTSC)|:|HD 720p@50Hz|:|HD [email protected]|:|HD 1080i@25Hz|:|HD [email protected]

Once I read in that line I use

$pieces = explode("|:|", $line);

This part is for generating C++ code now I have an array called variables, where I store name of the variable for C++ and what type it is.

$variables[] = array($pieces[1], $pieces[3]);

Once I loop through the text file and gather up all the data

I try to create a string which I will later on store inside a text file

$cpp .= "memset(prof,0,sizeof(prof));\n";
$cpp .= "sprintf(prof, \"CALL put_into_input(%d, 'test'";
for($i = 3; $i<count($variables); $i++)
{
    if(strpos($variables[0], "VARCHAR") === FALSE)
    {
        $cpp .= ", %d";
    }
    else
    {
        $cpp .= ", '%s'";
    }
}
$cpp .= "\nprofile-".">"."pfile";
for($i = 3; $i<count($variables); $i++)
{
    $cpp .= ",\n".$variables[1];
}
$cpp .= ");\n";

THE QUESTION

The problem is that when I try to concatenate string profile->input.videoStandard to my main string it simply saves it as Array, and I have noticed that php does not like -> as a string in general, which is why I used "-".">" wherever I could. How do I go about getting php of accepting -> as a string?

Thanks to anyone for their help.

How do You concatenate the string?

Michas
– Michas

2011-11-02 13:18:03 +00:00
Commented Nov 2, 2011 at 13:18 — Michas
– Michas, Commented Nov 2, 2011 at 13:18
I use . to concatenate strings.

Quillion
– Quillion

2011-11-02 14:50:55 +00:00
Commented Nov 2, 2011 at 14:50 — Quillion
– Quillion, Commented Nov 2, 2011 at 14:50

PeeHaa · Accepted Answer · 2011-11-02 16:01:47Z

2

I have noticed that php does not like -> as a string in general.

Nothing wrong with a -> in a string.

Unless there is some $ in there too.

This will cause PHP to think you are trying to access a property of an object instance.

E.g. $object->something. You could just use single quotes in stead of double quotes to prevent PHP trying to parse the string. Other than that there shouldn't be any problem using -> in a string.

EDIT

You're code looks strange to me.

You are doing $variables[] = array($pieces[1], $pieces[3]);

which is the same as $variables = array(array($pieces[1], $pieces[3]));

So you only have $variables[0] and not $variables[1]

edited Nov 2, 2011 at 16:01

answered Nov 2, 2011 at 12:50

PeeHaa

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9 Comments

Quillion Over a year ago

Thank you very much, that has worked! But what about reading in the string from a textfile? I can not control whether it has double or single quotes. So how would I go around that? The string being read in is still being displayed as Array instead of profile->input.videoStandard

PeeHaa Over a year ago

@Quillion: Have you changed all your double quotes to single quotes around your string in your PHP file?

Quillion Over a year ago

Yes every single one besides newline character, so if I want a newline character I do this ');'."\n"; but other than that every single one is a single quote. But my string that is being faulty is stored inside $variables[1] which I read from the text file, and I concatenate it like this $cpp .= ','."\n".$variables[1]; could this be the problem?

PeeHaa Over a year ago

Can you do a var_dump($variables[1]) or print_r($variables[1]); and report the result?

Quillion Over a year ago

I did that. when I print the value of $pieces[3] it says profile->input.videoStandard When I do $variables[] = array($pieces[1], $pieces[3]); Then do print_r it shows that it is an array that has INT or VARCHAR for first value and "" for second. Does the string somehow dissapear?

|

Tim S. · Accepted Answer · 2011-11-02 13:05:00Z

1

In PHP, variables within double quotes are parsed.

$var = "hello";
$str = "This is a variable: $var";

// The value of $str will be "This is a variable: hello"

When you use single quotes variables are not parsed.

$var = "hello";
$str = 'This is a variable: $var'; // Please note the single quotes here

// The value of $str will be "This is a variable: $var"

I think in your case, the -> are recognised as variables because that's the syntax used for object-oriented programming to get both properties and methods.

$object->property;
$object->method();

Personally I use single quotes whenever I don't need my string to be parsed for variables or regular expressions.

Single quote strings

Double quote strings

answered Nov 2, 2011 at 13:05

Tim S.

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Comments

Nanocom · Accepted Answer · 2011-11-02 12:54:18Z

-2

You could use apostrophes intead of quotes maybe ?

answered Nov 2, 2011 at 12:54

Nanocom

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