为什么在PHP中访问数组元素会引发字符串数组错误？

Question

I've done some searching and I didn't find any posts that quite answered my question. I have a PHP array generated, for the sake of argument, with this code:

$i = 5;
for($i = 0; $i < $j; $i++) {
   $multiArray[0][$i] = $i;
   $multiArray[1][$i] = $i;
}

When I try to access it with:

for($i = 0; $i < $j; $i++) {
   echo "$multiArray[0][$i]";
   echo "$multiArray[1][$i]";
}

I get:

Notice: Array to string conversion on line 3

Notice: Array to string conversion on line 4

...x4

When I replace echo with printf("%d", $multiArray[0][$i]) then it prints fine. Why do I have to explicitly tell PHP that I'm asking for an int when the element I'm accessing is clearly an int (and PHP knows it, via var_dump() )? I'm not accessing the array, but an element within the array.

Thanks

Answer 1

Simple double quoted variable interpolation supports up to one nested element. In other words, "foo[0][1]" is interpreted as "{$foo[0]}[1]" . That means it tries to interpret the array $foo[0] as a string at that point to interpolate it into the string.

But using quotes here at all is entirely nonsensical. You don't want string interpolation, you just want to output a variable value:

echo $multiArray[0][$i];

Answer 2

just try this:

for($i = 0; $i < $j; $i++) {
 echo $multiArray[0][$i]; 
 echo $multiArray[1][$i]; 
}

Answer 3

Your code is parsing the string not the array, try to remove the quotes.The [] brackets after the array are considered as a string not the as a paremeter.Use the code below

<?php
$j = 5;
for($i = 0; $i < $j; $i++) {
   $multiArray[0][$i] = $i;
   $multiArray[1][$i] = $i;
}

for($i = 0; $i < $j; $i++) {

   echo $multiArray[1][$i];
}

Hope this helps you