Parse error: syntax error in PHP

Question

I try to store a string to a PHP variable. But it shows an error "Parse error: syntax error, unexpected '='" Here is the code

$page-title = "Home Page";

PHP variables can't contain a minus sign. See: http://php.net/manual/en/language.variables.basics.php (What PHP does here is: `variable - constant = "string";` and the equal sign is unexpected here) — Rizier123, Aug 29 '16 at 09:08
its showing Dash Error which you using in your Variable Try : $page_title = "Home Page"; — Dharmendra, Aug 29 '16 at 09:09

score 0 · Answer 1 · answered Aug 29 '16 at 09:09

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Do like this:

$page_title = "Home Page";

answered Aug 29 '16 at 09:09

Virb

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*Why* should OP do it like this? – Rizier123 Aug 29 '16 at 09:10
@Rizier123: What will be your answer? – Virb Aug 29 '16 at 09:11
If you just give OP some code he will never know what he did wrong, so an explanation why OP has to do it like this will improve your answer. – Rizier123 Aug 29 '16 at 09:14

Red · Answer 2 · 2016-08-29T09:12:59.130

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Replace $page-title = "Home Page";

with: $page_title = "Home Page";

The variable you used is wrong. PHP variables cant contain - signs.

edited Aug 29 '16 at 09:12

answered Aug 29 '16 at 09:10

Red

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*Why* should OP do it like this? – Rizier123 Aug 29 '16 at 09:10
You right, i should be more clear – Red Aug 29 '16 at 09:12

score 0 · Answer 3 · answered Aug 29 '16 at 09:10

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Your variable name is incorrect.

A variable name can only contain alpha-numeric characters and underscores (A-z, 0-9, and _ ).

answered Aug 29 '16 at 09:10

Devsi Odedra

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You should be write as comment, not by an answer. – Virb Aug 29 '16 at 09:14
PHP variables can contain a bit more than alpha-numeric characters, see: http://php.net/manual/en/language.variables.basics.php . But it is correct, that the OP's variable name is invalid due to the minus sign in it. – Rizier123 Aug 29 '16 at 09:16

score 0 · Answer 4 · answered Aug 29 '16 at 09:13

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You can use it either $pageTitle or $page_title as your variable name fore more detail refer php manual

answered Aug 29 '16 at 09:13

Mohamed Nizar

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Masroor_Shah · Answer 5 · 2016-08-29T09:27:52.640

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you have use - subtraction sign in declaring variable that's why it cause an error you can check here how to declare php variableshow to declare php variables

try this

$page_title = "Home Page";

edited Aug 29 '16 at 09:27

answered Aug 29 '16 at 09:17

Masroor_Shah

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*Why* should OP "try this"? – Rizier123 Aug 29 '16 at 09:19
@Rizier123 coz - is used as arithmetic operator so it cause an error ................ Take a look here ,.....http://trainingphp.blogspot.com/2011/03/naming-rules-for-variables-variable.html ............................. – Masroor_Shah Aug 29 '16 at 09:23
@Rizier123 take a look here how to declare PHP variable http://www.w3schools.com/php/php_variables.asp – Masroor_Shah Aug 29 '16 at 09:24
Put that information inside your answer. – Rizier123 Aug 29 '16 at 09:25
@Rizier123 okay now so vote up me – Masroor_Shah Aug 29 '16 at 09:29

Parse error: syntax error in PHP

5 Answers5