打印变量和文本没有空格？

I have the following php code:

$mem=10;
echo "Memory: $mem MB<br>";

Which prints out:

Memory: 10 MB

I would like it to print out:

Memory: 10MB

I can't do this:

echo "Memory: $memMB<br>";

Any ideas how I can do this?

Using concatenation is an easy way:

echo "Memory: ".$mem."MB<br>";

echo "Memory: {$mem}MB<br>";

note that this work even with multi level arrays, or objects.

You can use commas with echo:

echo "Memory: ", $mem, "MB<br>";

Several ways, but IMHO, the best is:

printf('Memory: %dMB<br>', (int) $mem);//drop the cast if you so desire

That just looks clean, to me. In addition, using formatted output strings allows for a lot more control (ie: printf("Value: %.2f", 123.23213); will output Value: 123.23)
Alternatives would be:

echo 'Memory: ', $mem, '<br>';//comma's, because echo is a language construct
echo 'Memory: '.$mem.'<br>';//concatenate if you want

Or: use brackets to disambiguate the expression:

echo "Memory: {$mem}MB <br>";

Use concatenation,

echo 'Memory: '.$mem.'MB<br>';

Or else, this would also work,

echo "Memory: {$mem}MB<br>";

This is to separate your variable from the surrounding text.

<?php
    $mem=10;
    // you can do this because ther $mem is varibale 
    echo "Memory: $mem MB<br>";


    /*
        but here you cannot do this because Here $memMB act as a variable which 
        you not decleared give an error :  Notice: Undefined variable: memMB 
    */
    echo "Memory: $memMB<br>"; 

    /*
        For out put Memory: 10MB
    */
        echo "Memory:".$mem."MB"."<br />";
?>