PHP Parse Error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

Question

I've been trying to debug this for the past five minutes, I just don't get what the problem is:

Here's my code, lines 33 - 37:

for($i = 0; $i < 5; $i++) {
    $followers_change[$i] = $en_array1[$i]['followers']-$en_array2[$i]['followers'];
    $rank_change[$i] = $en_array1[$i]['rank']-$en_array2[$i]['rank'];
        echo "<tr><td>$en_array1[$i]['rank']</td><td><img src='$en_array1[$i]['imageurl']' width='48' height='48'/></td><td>$en_array1[$i]['name']</td><td>$en_array1[$i]['followers]'</td><td>$en_array1['followers_change']</td></tr>";
}

I keep getting the error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /.../ on line 36

I apologize for this being such a stupid/rookie error, I'm just stumped and tired at the same time (which doesn't make a good combination :)).

I personally avoid using array subscripts inside strings because it makes the kind of error you've got harder to detect. — zneak
– zneak, Commented Aug 31, 2011 at 2:48
String interpolation in double quotes follows specific rules. Use the curly braces syntax when in doubt. — mario
– mario, Commented Aug 31, 2011 at 3:29

Book Of Zeus · Accepted Answer · 2011-08-31 02:51:07Z

7

error in the followers key, and you should use this code:

  for($i = 0; $i < 5; $i++) {
      $followers_change[$i] = $en_array1[$i]['followers']-$en_array2[$i]['followers'];
      $rank_change[$i] = $en_array1[$i]['rank']-$en_array2[$i]['rank'];
          echo "<tr><td>" . $en_array1[$i]['rank'] . "</td><td><img src='" . $en_array1[$i]['imageurl'] . "' width='48' height='48'/></td><td>" . $en_array1[$i]['name'] . "</td><td>" . $en_array1[$i]['followers'] . "'</td><td>" . $en_array1['followers_change'] . "</td></tr>";
  }

edited Aug 31, 2011 at 2:51

answered Aug 31, 2011 at 2:41

Book Of Zeus

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5 Comments

Book Of Zeus Over a year ago

$en_array1[$i]['followers] it not valid

Avicinnian Over a year ago

Thanks, failure to properly encapsulate + the followers key issue was the problem :)! I didn't realize that I'd have to encapsulate a variable to be honest, as there's times when a string seems to echo fine inclusive of variables without encapsulation, and no it seems times when it doesn't. Any idea why the latter might have occurred? Would it be because of src='$en_array1[$i]['followers']'?

Book Of Zeus Over a year ago

You welcome, that is a good question. You are probably right, I'll have to dig into it to find why.

Tech4Wilco Over a year ago

Why not using the echo "<tr><td>{$en_array1[$i]['rank']}...?

Book Of Zeus Over a year ago

It's a personal preference, I think it's easier to read when you use the quotes and dots.

J-16 SDiZ · Accepted Answer · 2011-08-31 02:42:11Z

4

See :

 .... ><td>$en_array1[$i]['followers]'</td ...

Do you see the mismatched ' after followers] ?

answered Aug 31, 2011 at 2:42

J-16 SDiZ

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Comments

profitphp · Accepted Answer · 2011-08-31 02:42:50Z

4

<?

for($i = 0; $i < 5; $i++) {
    $followers_change[$i] = $en_array1[$i]['followers']-$en_array2[$i]['followers'];
    $rank_change[$i] = $en_array1[$i]['rank']-$en_array2[$i]['rank'];
        echo "<tr><td>{$en_array1[$i]['rank']}</td><td><img src='{$en_array1[$i]['imageurl']}' width='48' height='48'/></td><td>{$en_array1[$i]['name']}</td><td>{$en_array1[$i]['followers']}</td><td>{$en_array1['followers_change']}</td></tr>";
}

You're missing a ' and you should use {} around your vars

answered Aug 31, 2011 at 2:42

profitphp

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Comments

kungphu · Accepted Answer · 2011-08-31 02:41:16Z

3

You should be either using {} to encapsulate your variables in that string or using . concatenation.

answered Aug 31, 2011 at 2:41

kungphu

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