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[英]How to to send and retrieve data from flickr flickr.test.echo method using JQuery Ajax REST?
[英]how to retrieve echo data from php to jquery 1 by 1
我有一个单击的JQuery,它将php查询发送到MySql,然后我想在JQuery上将数据一一发送回去。
但是我只知道如何将结果从php发送回整个JQuery。
我目前的JQuery:
$(function() {
$(".img_thumb_holder").on("click", function() {
$.ajax({
type: "POST",
url: "CMS/PHP/retrieveAuthorDetails.php",
success: function(data) {
alert(data);
}
});
});
});
我目前的php:
<?php
include 'dbAuthen.php';
$sql = "SELECT * FROM users WHERE Name = 'james'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
echo $row['UserID'];;
echo $row['EmailAddr'];
}
?>
输出都是UserID和EmailAddr,我不知道如何只显示UserID或EmailAddr
我尝试过alert(data [0]),但是它只显示结果的一个字母。有关如何执行此操作的任何想法?
更新:在肖恩的帮助下,我有了当前更新的代码
jQuery的:
$(function() {
$(".img_thumb_holder").on("click", function() {
$.ajax({
type: "POST",
dataType: "json",
url: "CMS/PHP/retrieveAuthorDetails.php",
success: function(data) {
$.each(data, function() {
var userid = data.userid;
var useremail = data.email;
// i think there something wrong with this as it will keep repeating storing the userid and email for each data.. can someone verify?
});
}
});
});
});
PHP
<?php
include 'dbAuthen.php';
$sql = "SELECT * FROM users WHERE Name = 'honwenhonwen'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$arr = array(
"userid" => "HonWen",
"email" => "honwen@hotmail.com"
);
}
echo json_encode($arr);
?>
在您的php中,将结果保存到数组中-
<?php
include 'dbAuthen.php';
$array = array(); // create a blank array
$sql = "SELECT * FROM users WHERE Name = 'james'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
// add each result to the array
$array[] = array('UserID'=> $row['UserID'], 'EmailAddr'=> $row['EmailAddr']);
}
echo json_encode($array); // json_encode() the array
?>
然后在js / ajax中,您可以遍历每个值
$(function() {
$(".img_thumb_holder").on("click", function() {
$.ajax({
type: "POST",
url: "CMS/PHP/retrieveAuthorDetails.php",
success: function(data) {
// loop through each returned value
$.each(data, function(){
//alert each result, this is just an example as alert() for each result is not a great idea
alert("UserID:"+ this.UserID + " EmailAddr:" + this.EmailAddr);
});
}
});
});
});
jQuery的
$(function() {
$(".img_thumb_holder").on("click", function() {
$.ajax({
type: "POST",
dataType: "json",
url: "CMS/PHP/retrieveAuthorDetails.php",
success: function(data) {
$.each(data, function() {
var userid = data.userid;
var useremail = data.email;
// i think there something wrong with this as it will keep repeating storing the userid and email for each data.. can someone verify?
});
}
});
});
});
PHP
<?php
include 'dbAuthen.php';
$sql = "SELECT * FROM users WHERE Name = 'honwenhonwen'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
$arr = array(
"userid" => "HonWen",
"email" => "honwen@hotmail.com"
);
}
echo json_encode($arr);
?>
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