Question

我正在学习PHP并编写了一个简单的翻译器。

＆＃13;

<!DOCTYPE html>
<html lang="en">
<head>
	<meta charset="UTF-8">
	<title>Translator</title>
</head>
<body>

<form method="post" action="">
<input type="string" name="word">
<input type="submit">	
</form>

<?php 

if (isset($_POST["word"])) {
	$word = $_POST["word"];
	echo $word . " -> ";

function translate($word){
$dict = array('hello' => 'sawadee khap','thanks' => 'kap khum khap','sorry' => 'mai pen rai');
	foreach ($dict as $en => $th) {
		if ($word == $en) {
			echo $th;
			break;
		}
	}
} 

translate($word);

} else {
	echo "enter a word";
}

?>
</body>
</html>

＆＃13;

如何在词典中显示字符串＆＃39;当我输入一个不在数组中的单词？我也非常感谢有关改进代码的任何反馈或建议。

Answer 1

PHP有一个函数，称为in_array。你可以这样做：

$dict = array('hello' => 'sawadee khap','thanks' => 'kap khum khap','sorry' => 'mai pen rai');
if(!in_array($word, array_keys($dict))){
    echo '"' . $word . '" not found in the dictionary.';
}else{
    echo $dict[$word];
}

编辑：改进

$dict = array('hello' => 'sawadee khap','thanks' => 'kap khum khap','sorry' => 'mai pen rai');
if(!array_key_exists(strtolower($word), $dict)){
    echo '"' . $word . '" not found in the dictionary.';
}else{
    echo $dict[$word];
}

Answer 2

更改功能代码

function translate($word){
$dict = array('hello' => 'sawadee khap','thanks' => 'kap khum khap','sorry' => 'mai pen rai');

if(array_key_exists($word, $dict)){
     echo $dict[$word];
}else{
    echo 'not in dictionary';
}
}

Answer 3

如果在找到单词时从函数返回值，或者falsoe，否则您可以对结果进行逻辑测试以显示替代错误消息。

import traceback
import sys
from contextlib import contextmanager


@contextmanager
def output_to_file(filepath, write_mode='w'):
    stdout_orig = None
    stderr_orig = None

    stdout_orig = sys.stdout
    stderr_orig = sys.stderr

    f = open(filepath, write_mode)

    sys.stdout = f
    sys.stderr = f

    try:
        yield
    except:
        info = sys.exc_info()
        f.write('\n'.join(traceback.format_exception(*info)))

    f.close()

    sys.stdout = stdout_orig
    sys.stderr = stderr_orig

Answer 4

你可以使用array_key_exist：

$dict = array('hello' => 'sawadee khap','thanks' => 'kap khum khap','sorry' => 'mai pen rai');
if (array_key_exists($word, $dict)) {
//in dictionary
}else{
//not in dictionary
}

Answer 5

请试试这个

<form method="post" action="">
    <input type="string" name="word">
    <input type="submit">   
</form>

<?php

function translate($word) {
    $dict = array('hello' => 'sawadee khap', 'thanks' => 'kap khum khap', 'sorry' => 'mai pen rai');

    if (array_key_exists($word, $dict)) {
       echo $dict[$word];

    } else {
        echo " not in dictionary";
    }
}

if (isset($_POST["word"])) {
    $word = $_POST["word"];
    echo $word . " -> ";
    translate($word);
} else {
    echo "enter a word";
}
?>

Answer 6

你的top if语句缺少一个大括号。

关于你的问题，我假设你想要浏览字典，如果没有找到这个词来回应它。你可以使用一面旗帜。将它设置为for循环之前的某个内容，然后当您找到该单词时将其设置为其他内容。然后在for循环结束时检查，例如：

$found = false;
foreach ($dict as $en => $th) {
    if ($word == $en) {
        $found = true'
        echo $th;
        break;
    }
}
if (!$found) echo $word;''

PHP中的其他声明

6 个答案: