ตามหัวข้อเลยค่ะ ควรแก้ยังไงดีคะ โค้ดที่เขียนก็ตามนี้เลยค่ะ
<?php
header(\"content-type:text/javascript;charset=utf-8\");
define(\'HOST\',\'localhost\');
define(\'USER\',\'root\');
define(\'PASS\',\'\');
define(\'DB\', \'example\');
if($_SERVER[\'REQUEST_METHOD\'] == \'GET\'){
$status = $_GET[\'status\'];
$con = mysqli_connect(HOST,USER,PASS,DB) or die (\'ไม่สามารถเชื่อมต่อได้\')
mysqli_query(\"SET character_set_results=utf8\");
$sql = \"SELECT * FROM connect WHERE status=\'\" . $status . \"\'\" ;
$r = mysqli_query($con,$sql);
$result = array();
while($row = mysqli_fetch_array($r)){
array_push($result,array(\"comment\" => $row [\'comment\']));
}
echo json_encode(array(\'result\' => $result));
mysqli_close($con);
}
ขึ้น error ว่า syntax error, unexpected identifier "mysqli_query" แก้ยังไงดีคะ
<?php
header(\"content-type:text/javascript;charset=utf-8\");
define(\'HOST\',\'localhost\');
define(\'USER\',\'root\');
define(\'PASS\',\'\');
define(\'DB\', \'example\');
if($_SERVER[\'REQUEST_METHOD\'] == \'GET\'){
$status = $_GET[\'status\'];
$con = mysqli_connect(HOST,USER,PASS,DB) or die (\'ไม่สามารถเชื่อมต่อได้\')
mysqli_query(\"SET character_set_results=utf8\");
$sql = \"SELECT * FROM connect WHERE status=\'\" . $status . \"\'\" ;
$r = mysqli_query($con,$sql);
$result = array();
while($row = mysqli_fetch_array($r)){
array_push($result,array(\"comment\" => $row [\'comment\']));
}
echo json_encode(array(\'result\' => $result));
mysqli_close($con);
}