像在Java中一样将PHP对象传递到数组/列表中


passing php object into array/list like in java

也许它是重复的,但我找不到它。

    something = new ArrayList<>(); 
something.add(new Object("Hello"));
something.add(new object("World"));
something.add(new Object("!"));
for(blablabla){
System.out.print(something.get(i).getTextFromConstructor());
}

这将打印"你好世界!

在 PHP 中,我不知道将

整个对象传递到数组中以从循环或仅通过某些东西调用它们的方法的解决方案[0]->method();正如我知道这在 php 中无法完成,但也许我错了:-)

谢谢

$arr = array();
$arr[] = new MyUSerDefinedObject("Hello");
//...
echo $arr[0]->methd();
//or
foreach ($arr as $val) {
    echo $val->methd();
}

使用此示例设置桥接 b/w php 和 java,以便您可以将值传递给它们

<?php
$date = new Java("java.util.Date", 70, 9, 4);
var_dump($date->toString());
$map = new Java("java.util.HashMap");
$map->put("title", "Java Bridge!");
$map->put("when", $date);
echo $map->get("when")->toString()."'n";
echo $map->get("title")."'n";
$array = array(1,2,3,4,5);
$map->put("stuff", $array);
var_dump($map->get("stuff"))."'n";
$system = new JavaClass("java.lang.System");
echo "OS: ".$system->getProperty("os.name")."'n";
$math = new JavaClass("java.lang.Math");
echo "PI: ".$math->PI."'n";
?>