我有一个简单的文本区域，文字如下：
Lorem Ipsum [text] dolar, rock n'[more] roller.
我正在尝试检测光标在括号之间的时间，如果是这样，允许按Ctrl +向右或向左箭头将文本和括号向右或向左移动一个位置，每按一次按键，而不会超过开始或结束线，没有移过相邻的括号块。我猜这与将右侧第一个字符（或空格）复制到左侧的Ctrl +右箭头相同，反之亦然。
我有一些基本的jquery技能，但我对此的所有尝试都失败了。否则，我会粘贴一段显示我所拥有的代码。谢谢，
答案 0 :(得分：3)
解决方案需要几个不同的部分：
查找光标位置：
foreach (var row in rssDocument.SelectNodes("teryt/catalog/row"))
{
foreach (var col in row.SelectNodes("col"))
{
Console.WriteLine(col.InnerText);
}
}
找到字符串的位置（查看插入符号是否在其中）。您可以使用正则表达式，如：
function getCaretPosition(element)
{
var CaretPos = 0;
//Old IE way
if ( document.selection )
{
element.focus();
var textSelection = document.selection.createRange();
textSelection.moveStart( 'character', -element.value.length );
CaretPos = textSelection.text.length;
}
//DOM way
else if ( element.selectionStart || element.selectionStart == '0' )
CaretPos = element.selectionStart;
return (CaretPos);
}
您还需要在textarea和该事件监听器中收听按键事件：
这些作品应该使这项工作。 （我不打算为你写，因为那对你没有帮助）
答案 1 :(得分：0)
这是fiddle执行此操作，以下是代码：
$('textarea').on('keydown', function(e) {
// Is Ctrl-left and Ctrl+right pressed?
if (e.ctrlKey && (e.which === 37 || e.which === 39)) {
var pos = this.selectionStart;
var val = this.value;
var openBracketOnLeft = val.substr(0, pos).lastIndexOf('[');
var closeBracketOnLeft = val.substr(0, pos).lastIndexOf(']');
var closeBracketOnRight = val.substr(pos).indexOf(']');
// Is start of selection within two brackets?
if (openBracketOnLeft > closeBracketOnLeft && closeBracketOnRight !== -1) {
closeBracketOnRight += pos + 1;
var tagText = val.substr(openBracketOnLeft, closeBracketOnRight - openBracketOnLeft);
var level = 0;
// Repeat moving the tag until we do not break another tag in two.
do {
// Is Ctrl-left pressed, and is tag not yet on far left?
if (e.which === 37 && openBracketOnLeft) {
ch = val.substr(openBracketOnLeft - 1, 1);
val = val.substr(0, openBracketOnLeft - 1)
+ tagText
+ ch
+ val.substr(closeBracketOnRight);
openBracketOnLeft--;
closeBracketOnRight--;
// Is Ctrl-right pressed, and is tag not yet on far right?
} else if (e.which === 39 && closeBracketOnRight < val.length) {
ch = val.substr(closeBracketOnRight, 1);
val = val.substr(0, openBracketOnLeft)
+ ch
+ tagText
+ val.substr(closeBracketOnRight + 1);
openBracketOnLeft++;
closeBracketOnRight++;
} else {
break;
}
level += ch == '[' ? 1 : ch == ']' ? -1 : 0;
} while (level);
// Select the tag, without the brackets
this.value = val;
this.selectionStart = openBracketOnLeft + 1;
this.selectionEnd = closeBracketOnRight - 1;
e.preventDefault();
}
};
});
答案 2 :(得分：0)
这是一种没有邪恶的正则表达式字符串的方法。相反，我想尝试使用jQuery＆＃39; keydown＆＃39;与提问者提到的内容一致的事件（参见：jQuery中的newb）。还要注意＆＃39; keydown＆＃39;对于这种方法更好，因为关键词＆＃39;会发射多次，不过我想这也会发生......不管怎样，这就是我想出的：
$('#inputFieldInQuestion').on('keydown', function (event) {
// if both the control key and left key are pushed
if (event.keyCode == 37 && event.ctrlKey) {
// grab the text from the input and caret position in the input box
var inputBoxText = $(this).val(),
currentCaretPosition = this.selectionStart
// loop through all the characters in the input box text
for (var i = 0; i < inputBoxText.length; i++) {
// if the current character is an open bracket start testing for the end
if (inputBoxText[i] === "[") {
for (var j = i + 1; j < inputBoxText.length; j++) {
// this means that there is another bracketed string in between the
// beginning and the current bracketed string
if (inputBoxText[j] === "[") { break }
// if instead we come to the end of the bracketed string will determine
// if the bounds make sense
else if (inputBoxText[j] === "]") {
// if the caret position is in the bounds that you have just created
// we continue the shift
if (currentCaretPosition > i && currentCaretPosition < j) {
// test as per the question if the bracketed string is adjascent
// to another bracketed string
if (inputBoxText[i - 1] !== "]") {
// if the bracketed text is all the way to the left of the
// input box
if (i > 0) {
// slice and dice the string and move things left by one
// character
var frontString = inputBoxText.substring(0, i),
stringToMove = inputBoxText.substring(i, j + 1),
endString = inputBoxText.substring(j + 1)
$(this).val(frontString.slice(0, i - 1) + stringToMove + frontString.slice(i - 1) + endString)
this.setSelectionRange(currentCaretPosition - 1, currentCaretPosition - 1); break
}
}
else { break }
}
}
}
}
}
// important so that the ctrl-left doesn't shift the cursor to the end of the word
return false;
}
// if both the control key and right key are pushed
else if (event.keyCode == 39 && event.ctrlKey) {
var inputBoxText = $(this).val(),
currentCaretPosition = this.selectionStart
for (var i = 0; i < inputBoxText.length; i++) {
if (inputBoxText[i] === "[") {
for (var j = i; j < inputBoxText.length; j++) {
if (inputBoxText[j] === "]") {
if (currentCaretPosition > i && currentCaretPosition < j) {
// test as per the question if the bracketed string is adjascent
// to another bracketed string
if (inputBoxText[j + 1] !== "[") {
// bracketed text is all the way to the right of the input box
if (inputBoxText.length - j > 1) {
var frontString = inputBoxText.substring(0, i),
stringToMove = inputBoxText.substring(i, j + 1),
endString = inputBoxText.substring(j + 1)
$(this).val(frontString + endString.slice(0, 1) + stringToMove + endString.slice(1))
this.setSelectionRange(currentCaretPosition + 1, currentCaretPosition + 1); break
}
}
else { break }
}
}
}
}
}
return false;
}
})
这可能是执行此操作最复杂的方法，但它确实起作用并且满足所提出的所有约束。由于我刚刚注意到这是标记的正则表达式，这可能是一个可怕的解决方案。让掏膛开始！
超级奖励：如果您有任何数量的＆＃34; []＆＃34;字符串中的对。