我无法访问json数据,因为它总是失败并给出错误 SyntaxError:JSON.parse:JSON数据的第1行第1列的意外字符 search.php输出json数据但scripts.js输出json.parse错误 的的script.js
// execute when the DOM is fully loaded
$(function() {
console.log("1");
$("#q").typeahead({
autoselect: true,
highlight: true,
minLength: 1
},
{
source: search,
templates: {
empty: "no places found yet",
suggestion: _.template("<p><%- subname %></p>")
}
});
// re-center map after place is selected from drop-down
$("#q").on("typeahead:selected", function(eventObject, suggestion, name) {
});
});
function search(query, cb)
{
// get places matching query (asynchronously)
var parameters = {
sub: query
};
$.getJSON("search.php", parameters)
.done(function(data, textStatus, jqXHR) {
cb(data);
})
.fail(function(jqXHR, textStatus, errorThrown) {
console.log(errorThrown.toString());
});
}
的search.php
<?php
require(__DIR__ . "/../includes/config.php");
$subjects = [];
$sub = $_GET["sub"]."%";
$sql = "SELECT * from subjects where subname LIKE '$sub'";
echo $sql;
if($rows = mysqli_query($con,$sql))
{
$row = mysqli_fetch_array($rows);
while($row){
$subjects[] = [
'subcode' =>$row["subcode"],
'subname' => $row["subname"],
'reg' => $row["reg"],
'courseid' =>$row["courseid"],
'branchid' => $row["branchid"],
'yrsem' => $row["yrsem"]
];
$row = mysqli_fetch_array($rows);
}
}
// output places as JSON (pretty-printed for debugging convenience)
header("Content-type: application/json");
print(json_encode($subjects, JSON_PRETTY_PRINT));
?>
答案 0 :(得分:1)
您的问题的解决方案是 JSON.stringify
您需要将数据输出转换为字符串,即
function validateJSON(str) {
try {
JSON.parse(str);
} catch (e) {
return false;
}
return true;
}
然后使用
验证您的数据{{1}}
然后你就可以得到你的结果了。要进行验证,您可以通过以下函数传递数据: -
{{1}}