String.replace（）返回不需要的字符串 [英] String.replace() returns unwanted string

问题描述

我正在处理一段代码，我要将字符串拆分成单个部分并替换它们。 我的代码的基本逻辑流程是，有一个包含公式的字符串。 LHS下面的数字，即1,2和3是不同物体的ID。一旦我拆分它们，我就会使用这些ID，获取相应的值，并将下面String中的ID替换为各自的值。我所拥有的字符串如下 -

I'm working on a piece of code where I've to split a string into individual parts and replace them. The basic logic flow of my code is, there's a string that contains a formula. The numbers below on the LHS, i.e 1, 2 and 3 are ids of different objects. Once I split them, I'd use these ids, get the respective value and replace the ids in the below String with its respective values. The string that I have is as follow -

String str = "(1+2+3)>100";

我使用以下代码分割字符串 -

I've used the following code for splitting the string -

String[] arraySplit = str.split("\\>|\\<|\\=");
String[] finalArray = arraySplit[0].split("\\(|\\)|\\+|\\-|\\*");
arraySplit[0] = arraySplit[0].replace(id,reading);

因此，在拆分字符串后，我会用值替换字符串，即字符串现在是，（30 + 45 + 50）> 100 其中30,45和50是相应的值。 （此字符串将在SpEL中用于评估公式）

So, after the string is split, I'd replace the string with the values, i.e the string would now be, (30+45+50)>100 where 30, 45 and 50 are the respective values. (this string would then be used in SpEL to evaluate the formula)

在用值替换字符串时，我得到以下内容回复 -

While replacing the string with the values, I'm getting the following response -

初始字符串 -

（1 + 2 + 3）> 100 其中1,2和3是ID。 id的各个值分别为30.0,45.0和50.0。

(1+2+3)>100 where 1, 2 and 3 are ids. Respective values of the ids are 30.0, 45.0 and 50.0.

使用for循环替换值后，字符串为 -

After replacing the value by using a for loop, the string is -

（50.00.0 + 45.0 + 50.0）> 100 。第一个值很奇怪。原因是在第三次迭代中，字符串看起来像这样 - （30.0 + 45.0 + 3）> 100。因此，在第三次迭代中替换字符串时，将所有3的实例替换为50.0 ，从而产生上述字符串。所以显然使用 String.replace（）在这里不是更好的选择。那么在这种情况下我应该使用什么方法呢？我也在使用SpEL，所以如果使用SpEL有更好的解决方案，那么它也很好。

(50.00.0+45.0+50.0)>100. The first value is quite weird. The reason for this is while in the third iteration, the string looked like this - (30.0+45.0+3)>100. So while replacing the string in the third iteration, it replaces all instances of 3 with 50.0 hence resulting in the above string. So apparently using String.replace() won't be a better option here. What method should I use in this case then? I'm also using SpEL, so if there's a better solution to this using SpEL then also its good.

这里，（1 + 2 + 3）> 100 只是一个例子。圆括号是公式的一部分，字符串也可以是（（1 + 2 + 3）*（5-2））> 100 。或者甚至可以改变公式的语法，目前公式是这样的 - （1 + 2 + 3）* 4> 100 但是如果通过改变公式语法有一点出路然后这也有帮助，例如用 - （{＃1} + {＃2} + {＃3}）*
{＃4}> 100 替换公式，在这种情况下我使用{＃作为变量得到变量并得到数字。

Here, (1+2+3)>100 is just an example. The round braces are part of a formula, and the string could also be as ((1+2+3)*(5-2))>100. Or even the syntax of the formula could be changed, currently the formula is as such - (1+2+3) * 4>100 but if there's a way out by changing the formula syntax a bit then that would also be helpful, e.g replacing the formula by - ({#1}+{#2}+{#3}) * {#4}>100, in this case I'd get the variable using {# as the variable and get the numbers.

我希望这部分是明确的。

I hope this part is clear.

推荐答案

你没错， String.replace 不适合这里的工作。

You're right, String.replace is not the right tool for the job here.

您要做的是标记化输入字符串，然后对标记进行操作。您可以选择在最后组合回一个字符串，但这取决于您打算如何处理公式。

What you want to do is to tokenize the input string, then operate on the tokens. Optionally you can combine back into a string at the end, but it depends on what you intend to do with the formula.

例如，您想要标记字符串（1 + 2 + 3）> 100进入数组 [（，1，+，2， +，3，），>，100] 。这只是在数字和符号之间的每个边界上分裂的问题。

For example, you want to tokenize the string "(1+2+3)>100" into the array ["(", "1", "+", "2", "+", "3", ")", ">", "100"]. This is simply a matter of splitting on every boundary between a digit and a symbol.

从这个表示中，用它们替换数组中的整个标记是一件简单的事情。相应的值，例如找到值1的数组索引，并将其替换为30.0等。然后你可以创建一个新的表达式字符串，只需将数组中的所有部分连接在一起。

From this representation, it's a simple matter to replace the entire tokens in the array with their corresponding values, e.g. find the array index that has value "1" and replace it with "30.0", etc. Then you can create a new expression string by just joining all the parts in the array back together.

如果您想更进一步，可以转换为表达式树，它看起来像这样：

If you want to go one step further, you can translate into an expression tree, which would look something like this:

OpGreaterThan
    OpPlus
        Symbol(1)
        OpPlus
            Symbol(2)
            Symbol(3)
    Literal(100)

这个树是通过将输入表达式（以标准中缀表示法呈现）转换为反向抛光表示法，然后构建一个树，其中每个节点是一个运算符，其子节点是操作数。这将允许您象征性地操纵和/或评估公式。

This tree is formed by transforming the input expression (which is rendered in standard infix notation) into reverse polish notation, then building a tree where each node is an operator and its children are operands. This would allow you to symbolically manipulate and/or evaluate the formula.

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