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Parse Error Syntax Error Unexpected T_string Expecting


I feel that I have a grasp on the basics of PHP, but the syntax is confusing me. Parse or Fatal errors being more common. Outlet w/3 neutrals, 3 hots, 1 ground? Plugin Author jwineman @jwineman 8 months, 1 week ago No worries Scott - glad you were able to resolve your issue! his comment is here

Continue the conversation on Twitter or in a comment. Objects in PHP Forum View Course » View Exercise 595 points Submitted by Chip Eyler over 3 years ago Parse error: syntax error, unexpected T_STRING, expecting T_VARIABLE

Parse Error Syntax Error Unexpected T_string Expecting

See also: Stack Overflow question checklist" – bwoebi, DevZer0, swapnesh, OcramiusIf this question can be reworded to fit the rules in the help center, please edit the question. No single quotation marks this time) –Eugene Jul 17 '13 at 15:10 add a comment| up vote 1 down vote Try &body=%20%20%3A%20%20

echo $user->name, '!'; This line seems fine. In which case the error message provides a hint: expecting ‘,’ or ‘;’. For now, bonus points if you can find the other error. name) { echo 'It's time to stop writting errors "; echo $user->name, '!'; Interpreting PHP Unexpected Character In Input: '\' (ascii=92) State=1 Please use our new forums at discuss.codecademy.com.

Forum New Posts FAQ Calendar Forum Actions Mark Forums Read Quick Links Today's Posts View Site Leaders What's New? Parse Error: Syntax Error, Unexpected T_string Wordpress the sum of consecutive odd numbers Understanding the grammar: «illis Evangelii nuntiandi praebens mandatum» measurable linear functionals are also continuous on separable Banach spaces? Change syntax of macro, to go inside braces Amplifier circuit woe What are the ground and flight requirements for high performance endorsement? This problem is observed when an apostrophe sign is added in between a code that has been delimited with apostrophes, generating huge conflict with the PHP document.

echo 'we don\t know'; ?> Example 2 The Parse Error can also arise if the programmer has forgotten to insert the ; symbol at the end of the command.

if ($tried && $validated) {
echo '

The item has been created.


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