Question

我一直在玩array_search。我一直在尝试搜索数组并从中获取详细信息。问题是，一旦发现它继续运行的结果，它就不会检查数组中的重复项。我似乎无法解决我的循环逻辑，有什么建议吗？

$names数组

Array(
    [0] => Array
        (
            [name] => Fred
            [age] => 20
        )
    [1] => Array
        (
            [name] => George
            [age] => 17
        )
    [2] => Array
        (
            [name] => Sarah
            [age] => 22
        )
    [3] => Array
        (
            [name] => Fred
            [age] => 30
        )
)
$name=" Fred";
//loop here
$array_position = array_search($name, array_column($names, 'name'));
$array_data = $names[$array_position];

echo $array_data['name'] . " is: " . $array_data['age'] . " years old.
//end loop

结果：

弗雷德已经20岁了。

预期结果：

弗雷德已经20岁了

弗雷德今年30岁。

我试过array_keys()但我不太了解php.net

我应该尝试将结果存储在一个单独的数组中然后循环打印吗？似乎应该有一个更有效/更优雅的解决方案。

Answer 1

使用array_filter查找全部。

function search($array, $key)
{
  return array_filter($array, function($v) use($key){return $v['name'] == $key;});
}

$result = search($array, $key);
foreach($result as $v)
{
  echo $v['name'] . " is: " . $v['age'] . " years old.";
}

Answer 2

简单的foreach循环有什么问题？

// Declare the array
$names = array(
    array(
        'name' => 'Fred',
        'age' => 20
    ),
    array(
        'name' => 'George',
        'age' => 17
    ),
    array(
        'name' => 'Sarah',
        'age' => 22
    ),
    array(
        'name' => 'Fred',
        'age' => 30
    )
);

// Name to search
$name = 'Fred';

foreach($names as $index => $data){
    if($data['name'] == $name)
        echo $data['name'] . ' is ' . $data['age'] . ' years old.<br>';
}

结果将是：

Fred is 20 years old.
Fred is 30 years old.

Answer 3

array_keys()可以做到这一点：

<?php
$names = array(
    array(
        'name' => 'Fred',
        'age' => 20
    ),
    array(
        'name' => 'George',
        'age' => 17
    ),
    array(
        'name' => 'Sarah',
        'age' => 22
    ),
    array(
        'name' => 'Fred',
        'age' => 30
    )
);

$matchKeys = array_keys( array_column($names, 'name'), 'Fred' );
array_map(function($arrKey) use ($names) {
    echo $names[$arrKey]['name'] . " is: " . $names[$arrKey]['age'] . " years old.\n";
}, $matchKeys);

结果：

Fred is: 20 years old.
Fred is: 30 years old.

Answer 4

@Ben首先是array_search（）返回true或false而不是键，但是对于你想要的输出，你可以简单地尝试使用foreach（），条件如下或者使用array_key：

使用foreach（）

<?php
    $names = array(0 => 
                array(              
                        "name" => "Fred",
                        "age" => 20
                    ),
                1 => array(
                        "name" => "George",
                        "age" => 17
                    ),
                2 => array(
                        "name" => "Sarah",
                        "age" => 22
                    ),
                3 => array(
                        "name" => "Fred",
                        "age" => 30
                    )
            );

    $name= "Fred";
    echo "<pre>";
    foreach($names as $val){
        if($val["name"] == $name){
            echo $name . " is: " . $val["age"] . " years old.\n";
        }
    }

with array_key（）

<?php
    $names = array(0 => 
                array(              
                        "name" => "Fred",
                        "age" => 20
                    ),
                1 => array(
                        "name" => "George",
                        "age" => 17
                    ),
                2 => array(
                        "name" => "Sarah",
                        "age" => 22
                    ),
                3 => array(
                        "name" => "Fred",
                        "age" => 30
                    )
            );
    $name=" Fred";
    $array_position = array_keys(array_column($names, "name"), "Fred");
    foreach($array_position as $key){
        echo $names[$key]["name"] . " is: " . $names[$key]["age"] . " years old.\n";
    }

输出：

弗雷德是：20岁。

弗雷德是：30岁。

Answer 5

Array(
    [0] => Array
        (
            [name] => Fred,
            [age] => 20
        ),
    [1] => Array
        (
            [name] => George,
            [age] => 17
        ),
    [2] => Array
        (
            [name] => Sarah,
            [age] => 22
        ),
    [3] => Array
        (
            [name] => Fred,
            [age] => 30
        )
)

$name="Fred";

//loop here
for($i =0; i<count($names); $i++){
    if($arr[$i] == $name)echo $name . ' is ' . $names[$i] ."years old.<br>";
}
//end loop

输出：

Fred is 20 years old

Fred is 30 years old

Answer 6

这可以提供帮助 -

$check_name= 'Fred'; // set the name to check
$temp= array_filter($names, function($a) use($check_name) {
    return $a['name'] === $check_name; // return if matches
});
// loop through the result
foreach($temp as $v) {
   echo $v['name'] . ' is ' . $v['age'] . ' years old.<br>'; 
}

<强>输出

Fred is 20 years old.
Fred is 30 years old.

Working code

Anonymous function & use keyword

Answer 7

我不明白为什么你在$ name变量中给了空格 $ name =＆＃34;佛瑞德＆＃34 ;;

这是你的代码，

$name=" Fred";
function array_search_custom($needle,$haystack) {
    $temp = [];
    foreach($haystack as $key=>$value) {
        if(trim($needle)===trim($value)) {
            $temp[] =$key;

        }
    }
    return $temp;
}
$name=" Fred";
$a_names = array_column($names, 'name');
$a_ages = array_column($names, 'age');
$array_position = array_search_custom($name, $a_names);
foreach($a_names as $k => $v){
  if($array_position && in_array($k, $array_position)){    
    echo $a_names[$k] . " is: " . $a_ages[$k] . " years old.<br/>";
  }
}

我修剪了你的名字并根据你的要求编写了自定义数组搜索功能。

尝试一下，它会起作用。

使用array_search但在找到结果后继续搜索（PHP）

7 个答案: