我有一個(gè)我無法理解的情況。例如，我有這個(gè)關(guān)聯(lián)數(shù)組$assocHeroes = array( array( "name" => "Markos", "type" => "boy", ), array( "name" => "Jim", "type" => "boy", ), array( "name" => "Maria", "type" => "liz", ));$wallOfHeroes = array();foreach ($assocHeroes as $assocHero){ foreach ($assocHero as $name=>$type){ $hero = new Hero($name,$type); echo $hero->taunt(); $wallOfHeroes[]=$hero; }}print_r($wallOfHeroes);正如您從代碼中看到的那樣，$hero = new Hero($name,$type);我還有一個(gè) Class Hero。在這個(gè)類中，我只有一個(gè)功能taunt();，更具體地說public function taunt() { return "My name is {$this->name}!".PHP_EOL;}所以我只想得到My name is Markos My name is Jim..等..我不知道我做錯(cuò)了什么
2 回答
哈士奇WWW
TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超6個(gè)贊
首先，由于您沒有給出類的代碼以及如何從構(gòu)造函數(shù)中分配值，因此很難告訴您哪里錯(cuò)了。
我已根據(jù)您的問題大綱自由定義課程并附上一組工作代碼。這應(yīng)該工作。
<?php
class Hero
{
public $_name;
public $_type;
public function __construct($name,$type)
{
$this->_name = $name;
$this->_type = $type;
}
public function taunt()
{
return "My name is ".$this->_name."!<br>";
}
}
$assocHeroes = array(
array(
"name" => "Markos",
"type" => "boy",
),
array(
"name" => "Jim",
"type" => "boy",
),
array(
"name" => "Maria",
"type" => "liz",
)
);
$wallOfHeroes = array();
for($i=0;$i<count($assocHeroes);$i++)
{
$hero = new Hero($assocHeroes[$i]["name"],$assocHeroes[$i]["type"]);
echo $hero->taunt();
}
?>
慕沐林林
TA貢獻(xiàn)2016條經(jīng)驗(yàn) 獲得超9個(gè)贊
你可以試試foreach，確保值可以打印，用于調(diào)試你的代碼
foreach ($assocHeroes as $assocHero){
foreach ($assocHero as $key => $value){
echo $key ." ". $value . "<br>";
}
}
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