Javascript json_encode json.parse意外标记<;
我的代码:Javascript json_encode json.parse意外标记<;,javascript,json,Javascript,Json,我的代码: function callback(request, prim_key) { var type = request.getResponseHeader("Content-Type"); if (type === "application/json") { console.log(request.responseText); var response = JSON.parse(request.responseText);
function callback(request, prim_key) {
var type = request.getResponseHeader("Content-Type");
if (type === "application/json") {
console.log(request.responseText);
var response = JSON.parse(request.responseText);
if (!response['success']) {
alert("Error!\n\nError number: " + response['errno'] + "\n\nError string: " + response['error']);
}
else {
alert("Update Successful!");
}
var els = getElementsByClass(prim_key);
for (var i = 0; i < els.length; i++) {
if (els[i].type != 'checkbox') {
els[i].innerHTML = response[els[i].headers];
}
}
} else {
alert("not a json reponse! check database for errors.");
}
}
似乎没有一个可能的原因是您在
php.ini
中启用了ob\u start(“ob\u tidyhandler”)
。如果是,则注释此行以禁用它
似乎是原因。如果您检查修改后的缓冲区,它将返回与您发布的结果的标题完全相同的结果
希望这有帮助。发布JSON,但它可能是一个javscript对象。您可以使用Chrome或Firefox的控制台设置断点,并查看JSON的情况。对于abc123的评论,您确定您也没有发送回HTML吗?如果它不能解析第一个字符,这将是一个“When you said”获取下面的结果:“我想你忘了复制结果,所以我们不知道它是什么。当我得到这个结果时,通常是因为服务器端脚本(PHP、ASP等)出错。因此,如果您是通过服务器端语言检索JSON,请检查。Chirstopher,您是对的,我忘记复制结果了。请参阅我的更新问题,谢谢。我的php.ini中没有ob_start(“ob_tidyhandler”)。我应该在哪里找到它?谢谢。按照你的想法,我加上“ob_end_clean();”在我的php文件中的“header和json_encode($request)”之前,它现在可以工作了。谢谢
{"success":true,"ID":"131","Manufacturing_ID":"88888","Sequence_Name":"NDU","Bases":"22","Sequence":"TGG AGA TGT GTG CGT TCT TCTGG AGA TGT GTG CGT TCT TCTGG AGA TGT GTG CGT TCT TC","Tm_C":"55.00000000","nmoles":"26.0","Owner":"Liang","Species":"gallus"}
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"...................
{"success":true,"ID":"131","Manufacturing_ID":"88888","Sequence_Name":"NDU","Bases":"22","Sequence":"TGG AGA TGT GTG CGT TCT TCTGG AGA TGT GTG CGT TCT TCTGG AGA TGT GTG CGT TCT TC","Tm_C":"55.00000000","nmoles":"26.0","Owner":"Liang","Species":"gallus"}
$query = "SELECT * FROM $table WHERE ";
$i = 0;
foreach ($pks as $pk) {
$query .= "$pk='$pvs[$i]' AND ";
$i++;
}
$query = rtrim($query, " AND ");
$result = $db_conn->query($query);
$row = $result->fetch_assoc();
//$flag_key='';
if ($db_conn->errno) {
echo "oops";
} else {
foreach ($row as $key => $value) {
$response[$key] = $value;
//$flag_key .=$key.'-'.$value.'***';
}
}
header('Content-Type: application/json');
echo json_encode($response);