您正在使用$result[]['id']=2/1;
构建嵌套数组。每个调用都将一个带有单个元素的新关联数组添加到$result
列表数组。
PHP将两件事编码为JSON对象-关联数组和对象。使用数组的时间较短,但是使用对象可以帮助更好地理解数组。因此,首先使用stdClass
的示例:
$json = new stdClass();
$json->data = $data = new stdClass();
$data->id = 2;
$data->email = "janet.weaver@reqres.in";
$data->first_name = "Janet";
$data->last_name = "Weaver";
$data->avatar= "some url";
echo json_encode($json,JSON_PRETTY_PRINT);
对象是PHP中的引用。因此,更改将影响引用的所有变量。数组不是这种情况。但这可以只用一个表达式来完成:
$json = [
'data' => [
'id' => 2/1,'email' => "janet.weaver@reqres.in",'first_name' => "Janet",'last_name' => "Weaver",'avatar' => "some url"
]
];
echo json_encode($json,JSON_PRETTY_PRINT);
,
希望此示例对您有所帮助,以下示例具有您期望的所有组合。
<?php
$name = "Your Name";
$yearsofexp = "Years";
$designation = "Your designnation";
$pro_summary->name = $name;
$pro_summary->yearsofexp = $yearsofexp;
$pro_summary->designation = $designation;
$topics->tech1 = "Android";
$topics->tech2 = "Java";
$topics->tech3 ="PHP";
$past_exp[]=array("comp"=> "company One","role"=> "techlead","dates"=>"2015-2019","main_resp"=>"coding,testing,debugging","logo"=>"some.png");
$past_exp[]=array("comp"=> "company 2","role"=> "senior software engineer","dates"=>"2014-2015","logo"=>"some2.png");
$past_exp[]=array("comp"=> "company 3","role"=> "senior software developer","dates"=>"2013-2014","logo"=>"some3.png");
$past_exp[]=array("comp"=> "company 4","role"=> "software developer","dates"=>"2010-2013","logo"=>"some4.png");
//$past_exp[] = array("px1" =>$past_exp[1],"px2" =>$past_exp[2],"px3" =>$past_exp[3],"px4" => $past_exp[4] );
$uni[] = array("education"=>"Degree 1","university"=>"University 1","logo" =>"uni1.png");
$uni[] = array("education"=>"Degree 2","university"=>"University 2","logo"=>"uni2.png");
$my_cv->Professional = $pro_summary;
$my_cv->Technology = $topics;
$my_cv->PastExperience = $past_exp;
$my_cv->Education = $uni;
echo json_encode($my_cv);
?>
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