Error declaring variable php

Question

I want to Declaring variable

and I get error

Parse error: syntax error, unexpected 'book' (T_STRING)

mycode

$filters = "library&loc=local,scope:("book")&loc=local,scope:("book2")....";
$path = "http://xxxxx/place/search/institution=".$filters."";

in line $filters=..... is error

and I don't want to change "..." to '...' because my result is change

Look at the highlighting. You'll need to escape the quotes if you don't want to change to singlequotes, making it `...scope:(\"book\")&...` — Qirel, Mar 08 '17 at 14:20

Jonathon · Accepted Answer · 2017-03-08T14:34:44.297

3

You need to escape the double quotes:

$filters = "library&loc=local,scope:(\"book\")&loc=local,scope:(\"book2\")....";

Alternatively, you can use single quotes for the entire string:

$filters = 'library&loc=local,scope:("book")&loc=local,scope:("book2")....';

Both of these result in the same string which is library&loc=local,scope:("book")&loc=local,scope:("book2")....

edited Mar 08 '17 at 14:34

answered Mar 08 '17 at 14:21

Jonathon

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score 1 · Answer 2 · answered Mar 08 '17 at 14:22

1

You are using double quotes in the variable wich caused the error. You need to escape these quotes for example:

scope:(\"book\")

Or use single quotes:

scope:('book')

Same for book2.

I hope this will help!

answered Mar 08 '17 at 14:22

node_modules

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score 1 · Answer 3 · answered Mar 08 '17 at 14:22

1

@ARR.s use \before " like \" try below one:

<?php
$filters = "library&loc=local,scope:(\"book\")&loc=local,scope:(\"book2\")....";
$path = "http://xxxxx/place/search/institution=".$filters."";

answered Mar 08 '17 at 14:22

lazyCoder

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score 1 · Answer 4 · answered Mar 08 '17 at 14:22

1

This should fix it. Use single quotes at the start and end of your string:

$filters = 'library&loc=local,scope:("book")&loc=local,scope:("book2")....';

answered Mar 08 '17 at 14:22

Ramy Talal

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score 1 · Answer 5 · answered Mar 08 '17 at 14:24

You're using wrong the double quotes:

$filters = "library&loc=local,scope:("book")&loc=local,scope:("book2")....";

If you want to use double quotes inside a string declared using double quotes you have to scape them:

$filters = "library&loc=local,scope:(\"book\")&loc=local,scope:(\"book2\")....";

Anyway, the difference between using double and single quotes in php is that double quotes are evaluated, in your case you're not evaluating anything, so you can do this (and is a little faster becouse php doesn't have to evaluate the string)

$filters = 'library&loc=local,scope:("book")&loc=local,scope:("book2")....';

I recommend you to read the php strings documentation to understand the difference between double quoted and single quoted strings and how string evaluation works

Error declaring variable php

5 Answers5