我需要从 php 数据中提取 URL,我该如何实现?
.PHP
$query = 'SELECT * FROM picture LIMIT 3';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url.=$rec['pic_location'].";";
}
echo json_encode($url);
阿贾克斯
<script type="text/javascript">
$(document).ready(function() {
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first: " + arr[0] + ", second: " + arr[1]);
alert(arr[0]);
$(".theImage").attr("src", +arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
警报消息不起作用,它只是打印 H T(我认为这些是 http://...
您返回的字符串未解析为 JSON。只需将dataType: "json"
添加到 ajax 设置中即可。
由于你在javascript中将其作为数组读取,因此您应该像这样返回它:
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
你在PHP中发送一个字符串,并期望一个数组作为javascript中的响应。将 PHP 更改为
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url[] = $rec['pic_location'];
}
和 JavaScript 到
$.ajax({
url: "viewnew.php",
dataType: "JSON",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret[0]);
var arr = ret;
alert(arr);
alert("first: " + arr[0] + ", second: " + arr[1]); // THIS IS NOT WORKING!!!!
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});