php 类和 jQuery

Question

我有简单的自动完成 JQuery

$(document).ready(function() {
$( ".autocomplete" ).each(function() {
    $(this).autocomplete({
        source: '/php/LiveSearch2.php?name='+ $(this).attr('name') + '&',
        minLength: 1,
        appendTo: "#container"
    }); 
});
});

我已经编写了 php 类来推送值，但它不起作用。我在某个地方犯了一些错误。下面是LiveSearch2.php

<?php

class swapnil {

private $testname;
public $json;

public function __construct {
    $this->testname = array('swapnil','ranadive');
    $this->json = json_encode($this->testname);
    echo $this->json;

}
}

$Business = new swapnil();
?>

当我使用没有类的 php 代码时，它工作正常。

<?php
 $testname = array('swapnil', 'ranadive');
 $json = json_encode($testname);

 echo $json
 ?>

Answer 1

__construct 之后缺少括号。这里试试这个:

 class swapnil {

 private $testname;
 public $json;

 public function __construct() {
  $this->testname = array('swapnil','ranadive');
  $this->json = json_encode($this->testname);
  echo $this->json;

 }
}

$Business = new swapnil();