Javascript 将数据从php传回ajax
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Passing data back from php to ajax
提问by Yahoo
How can I pass data from a php of then rows back to ajax ?
如何将数据从 php 的行传回 ajax ？
PHP
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[]=$rec['pic_location'];
$name[]=$rec['name'];
$age[]=$rec['age'];
$gender[]=$rec['gender'];
}
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Ajax
阿贾克斯
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
dataType: "json",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first image url: " + arr[0][0] + ", second image url: " + arr[0][1]); // This code isnt working
alert("first image Name: " + arr[1][0] + ", second image name: " + arr[1][1]);
$(".theImage").attr("src", arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
My question is how can I display the values here ? The Alert message is giving me "Undefined" ?
我的问题是如何在此处显示值？警报消息给我“未定义”？
回答by Alexander
You can do something along these lines.
你可以沿着这些方向做一些事情。
PHP
PHP
$query = 'SELECT * FROM picture order by rand() LIMIT 10';
$res = mysql_query($query);
$pictures = array();
while ($row = mysql_fetch_array($res)) {
$picture = array(
"pic_location" => $row['pic_location'],
"name" => $row['name'],
"age" => $row['age'],
"gender" => $row['gender']
);
$pictures[] = $picture;
}
echo json_encode($pictures);
JS
JS
...
$.ajax({
...
dataType: "json",
...
success: function(pictures){
$.each(pictures, function(idx, picture){
// picture.pic_location
// picture.name
// picture.age
// picture.gender
});
}
});
...
回答by Sarfraz
You can't put multiple echostatements for the AJAX response:
您不能echo为 AJAX 响应放置多个语句：
echo json_encode($url);
echo json_encode($name);
echo json_encode($age);
echo json_encode($gender);
Join your arrays and send a single response:
加入您的数组并发送一个响应：
$arr = $url + $name + $age + $gender;
echo json_encode($arr);
回答by Srinivasan Annamalai
You can easily do this using a single Array:
您可以使用单个数组轻松完成此操作：
$pics = array();
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$pics[$rec['id']]['url'] = $rec['pic_location'];
$pics[$rec['id']]['name']=$rec['name'];
$pics[$rec['id']]['age']=$rec['age'];
$pics[$rec['id']]['gender']=$rec['gender'];
}
echo json_encode($pics);
