Php 返回JSON，使用Javascript编写
嗨 我有个小问题。我试图写一篇ajax文章来从数据库中获取值，它返回的格式是JSON对象。 如何从中获取键和值对 发送ajax的jquery：Php 返回JSON，使用Javascript编写,php,javascript,ajax,json,Php,Javascript,Ajax,Json,嗨 我有个小问题。我试图写一篇ajax文章来从数据库中获取值，它返回的格式是JSON对象。 如何从中获取键和值对 发送ajax的jquery： $ordering = array ("buy_type " . $_POST['buyType'],"prop_type ".$_POST['propertyType'],"district ".$_POST['disctrict'], "street ".$_POST['street'],"room_m
$ordering = array ("buy_type " . $_POST['buyType'],"prop_type ".$_POST['propertyType'],"district ".$_POST['disctrict'],
"street ".$_POST['street'],"room_min ".$_POST['roomMin'],"room_max ".$_POST['roomMax'],
"price_min ".$_POST['priceMin'],"price_max ".$_POST['priceMax'],"condition_type ".$_POST['conditionType'],"heat_type ".$_POST['heatType'],"lift_type ".$_POST['liftType'],"parking_type ".$_POST['parkingType']);
$user=$_SESSION["user"];
$whois = $mysqli->query('SELECT * FROM users WHERE uid='.$mysqli->real_escape_string($user).' ');
$who = $whois->fetch_assoc();
switch($who['user_title']){
case '0':
$res=$mysqli->query('SELECT * FROM searches WHERE uid='.$mysqli->real_escape_string($user).'
ORDER BY '.$mysqli->real_escape_string($ordering[0]).',
'.$mysqli->real_escape_string($ordering[1]).',
'.$mysqli->real_escape_string($ordering[2]).',
'.$mysqli->real_escape_string($ordering[3]).',
'.$mysqli->real_escape_string($ordering[4]).',
'.$mysqli->real_escape_string($ordering[5]).',
'.$mysqli->real_escape_string($ordering[6]).',
'.$mysqli->real_escape_string($ordering[7]).',
'.$mysqli->real_escape_string($ordering[8]).',
'.$mysqli->real_escape_string($ordering[9]).',
'.$mysqli->real_escape_string($ordering[10]).',
'.$mysqli->real_escape_string($ordering[11]).'
') or die($mysqli->error);
while($ki=$res->fetch_assoc()){
$tomb[] = $ki;
}
$tomb = array("query-data"=>$tomb);
echo json_encode($tomb);
Object {query-data: Array[3]}
query-data: Array[3]
0: Object
buy_type: "kiado"
condition_type: "uj"
district: "1"
heat_type: "cirko"
id: "1"
lift_type: "all"
parking_type: "all"
price_max: "22"
price_min: "10"
prop_type: "lakas"
room_max: "3"
room_min: "1"
street: "all"
uid: "3"
__proto__: Object
1: Object
2: Object
length: 3
__proto__: Array[0]
__proto__: Object
$ordering = array ("buy_type " . $_POST['buyType'],"prop_type ".$_POST['propertyType'],"district ".$_POST['disctrict'],
"street ".$_POST['street'],"room_min ".$_POST['roomMin'],"room_max ".$_POST['roomMax'],
"price_min ".$_POST['priceMin'],"price_max ".$_POST['priceMax'],"condition_type ".$_POST['conditionType'],"heat_type ".$_POST['heatType'],"lift_type ".$_POST['liftType'],"parking_type ".$_POST['parkingType']);
$user=$_SESSION["user"];
$whois = $mysqli->query('SELECT * FROM users WHERE uid='.$mysqli->real_escape_string($user).' ');
$who = $whois->fetch_assoc();
switch($who['user_title']){
case '0':
$res=$mysqli->query('SELECT * FROM searches WHERE uid='.$mysqli->real_escape_string($user).'
ORDER BY '.$mysqli->real_escape_string($ordering[0]).',
'.$mysqli->real_escape_string($ordering[1]).',
'.$mysqli->real_escape_string($ordering[2]).',
'.$mysqli->real_escape_string($ordering[3]).',
'.$mysqli->real_escape_string($ordering[4]).',
'.$mysqli->real_escape_string($ordering[5]).',
'.$mysqli->real_escape_string($ordering[6]).',
'.$mysqli->real_escape_string($ordering[7]).',
'.$mysqli->real_escape_string($ordering[8]).',
'.$mysqli->real_escape_string($ordering[9]).',
'.$mysqli->real_escape_string($ordering[10]).',
'.$mysqli->real_escape_string($ordering[11]).'
') or die($mysqli->error);
while($ki=$res->fetch_assoc()){
$tomb[] = $ki;
}
$tomb = array("query-data"=>$tomb);
echo json_encode($tomb);
有人能帮我把这些值写到一个表中吗？你可以通过这样一个对象进行循环： 没有从原型继承属性
for( i in json ) {
if ( json.hasOwnProperty( i ) ) {
console.log( i, json[ i ] );
}
}
for( i in json ) {
console.log( i, json[ i ] );
}
for( i in json ) {
if ( json.hasOwnProperty( i ) ) {
console.log( i, json[ i ] );
}
}
for( i in json ) {
console.log( i, json[ i ] );
}
ijson[i]$.post( 'loader.php', getGetStr(), function( data ) {
if ( !data || !data['query-data'] ) {
// invalid json string, so dont process
return;
}
data = data['query-data'];
// create the table
var table = $("<table />").html('<thead></thead><tbody></tbody>');
// inserted table head cols?
var thead = false;
// loop through 'query-data'
for( i = 0; i < data.length; i++ ) {
// append 'tr' element to 'tbody'
var tr1 = $("<tr />").appendTo( table.find("tbody") );
if ( !thead ) {
// if not finished creating table head cols, then append 'tr' elemnts to thead
var tr2 = $("<tr />").appendTo( table.find("thead") );
}
// loop if its an object
if ( typeof data[i] === "object" ) {
for( j in data[i] ) {
if ( !thead ) {
// if not finished creating table head cols, then append 'th' elements to thead
$("<th />").html( j ).appendTo( tr2 );
}
// insert our real dat to table rows
$("<td />").html( data[i][j] ).appendTo( tr1 );
}
// we finished creating table head cols
thead = true;
}
}
// append the table to whatever element you want,
// you can also use $("body").html( table );
table.appendTo( $("body") );
}, "json" );
$.post（'loader.php'，getstr（），函数（数据）{
如果（！data | |！data['query-data']））{
//无效的json字符串，请不要处理
返回；
}
数据=数据['query-data']；
//创建表
变量表=$（“”）.html（“”）；
//插入式桌头cols？
var thead=假；
//循环“查询数据”
对于（i=0；i编写要显示的动态html表格。以下是打印表格的原则
function printRow(item){
var html = "<tr>";
for(var key in item){
html += "<td>"+key+"</td><td>"+item[key]+"</td">;
}
html += "</tr>";
return html;
}
$.post('loader.php',getGetStr(),function(data){
var json = $.parseJSON(data);
console.log(json);
for(var key in json['query-data']){
var item = json['query-data'][key];
var html = "<table>";
html += printRow(item);
html += "</table>";
}
});
函数打印行（项）{
var html=“”；
用于（var输入项）{
html++++key++++item[key]+“您尝试过什么？它只是一个普通的Javascript对象，您可以使用json['query-data']
访问它。这是一个可以迭代的数组。只是一个建议，将一个额外的参数传递给$。post
作为json
发布，因此您不需要使用$。parseJSON
ie:$。post（url，params，callback，'json'）
；但是输出没有什么不同。我怎样才能得到像key，value这样的值呢？但是我怎样才能从这些查询中写入表呢？不要使用for..in
来迭代数组。