我想要成功函數來獲取它應該從PHP文件接收到的字符串。爲什麼我返回[Object,object]而不是字符串?
PHP抱怨「mysql_fetch_array()期望參數1是一個資源,而是給出布爾值」。我認爲這就是成功功能不會觸發的原因。
我在做什麼錯?
jQuery的:
var string = "something to be inserted";
$.ajax({
url: '...',
type: 'post',
dataType: 'json',
data: {toBeInserted: string.toLowerCase()},
success: function(data) {
console.log(data);
// some code that is to work with data
}
});
的PHP:
include 'serverStuff.php';
// A separate file with $con defined in it. Assume this works.
mysql_query("SET NAMES 'utf8'", $con);
// inserts the $_POST['toBeInserted'] into the database just fine
// assume the following are defined:
// (string) $user_name, (string) $now, (string) $statement
$sql=("SELECT * FROM table WHERE user_name=$user_name AND date=$now AND statement = $statement");
$result=mysql_query($sql, $con);
if ($row = mysql_fetch_array($result)) {
$new_id = (int) $row['id'];
}
mysql_close($con);
echo json_encode($new_id.'_'.$now);
嘗試'的console.log(JSON.stringify(數據));'相反,控制檯將打印對象作爲[對象,對象],如果你只是打印對象本身。 – yvesmancera
完成。這是日誌:{「readyState」:4,「responseText」:「\」67_18:52:54 21-8-2015 \「」,「status」:200,「statusText」:「OK」} – VHK