1 /*
2 * This file is part of the CSS implementation for KDE.
3 *
4 * Copyright 1999 Lars Knoll (knoll@kde.org)
5 *
6 * This library is free software; you can redistribute it and/or
7 * modify it under the terms of the GNU Library General Public
8 * License as published by the Free Software Foundation; either
9 * version 2 of the License, or (at your option) any later version.
10 *
11 * This library is distributed in the hope that it will be useful,
12 * but WITHOUT ANY WARRANTY; without even the implied warranty of
13 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
14 * Library General Public License for more details.
15 *
16 * You should have received a copy of the GNU Library General Public License
17 * along with this library; see the file COPYING.LIB. If not, write to
18 * the Free Software Foundation, Inc., 51 Franklin Street, Fifth Floor,
19 * Boston, MA 02110-1301, USA.
20 *
21 */
22 #include "csshelper.h"
23
24 #include "misc/helper.h"
25 #include "xml/dom_stringimpl.h"
26
27 using namespace DOM;
28 using namespace khtml;
29
parseURL(const DOMString & url)30 DOMString khtml::parseURL(const DOMString &url)
31 {
32 DOMStringImpl *i = url.implementation();
33 if (!i) {
34 return DOMString();
35 }
36
37 int o = 0;
38 int l = i->l;
39 while (o < l && (i->s[o] <= ' ')) {
40 o++;
41 l--;
42 }
43 while (l > 0 && (i->s[o + l - 1] <= ' ')) {
44 l--;
45 }
46
47 if (l >= 5 &&
48 (i->s[o].toLower() == 'u') &&
49 (i->s[o + 1].toLower() == 'r') &&
50 (i->s[o + 2].toLower() == 'l') &&
51 i->s[o + 3].toLatin1() == '(' &&
52 i->s[o + l - 1].toLatin1() == ')') {
53 o += 4;
54 l -= 5;
55 }
56
57 while (o < l && (i->s[o] <= ' ')) {
58 o++;
59 l--;
60 }
61 while (l > 0 && (i->s[o + l - 1] <= ' ')) {
62 l--;
63 }
64
65 if (l >= 2 && i->s[o] == i->s[o + l - 1] &&
66 (i->s[o].toLatin1() == '\'' || i->s[o].toLatin1() == '\"')) {
67 o++;
68 l -= 2;
69 }
70
71 while (o < l && (i->s[o] <= ' ')) {
72 o++;
73 l--;
74 }
75 while (l > 0 && (i->s[o + l - 1] <= ' ')) {
76 l--;
77 }
78
79 DOMStringImpl *j = new DOMStringImpl(i->s + o, l);
80
81 int nl = 0;
82 for (int k = o; k < o + l; k++)
83 if (i->s[k].unicode() > '\r') {
84 j->s[nl++] = i->s[k];
85 }
86
87 j->l = nl;
88
89 return j;
90 }
91