插入數據到數據庫並將其顯示到表中jquery onclick事件

Question

我的jQuery代碼在這裏有什麼問題嗎？ 我想插入數據並使用termo_i.php顯示數據。插入數據的腳本正在運行，但我認爲用於顯示數據庫值的腳本失敗。當你給我一個例子時，我會很感激。

$(document).on('click','#ok',function(e) { 

     if ($('#no').val() == '') { 
        alert('Kolom No Tolong Di Isi dan Tidak Boleh Sama Dimulai Dari 1'); 
       } else { 
     var data = $("#form_input").serialize(); 
     $.ajax({ 
     type: "Post", 
      url: "../php/termo_i.php", 
      success: function(data){ 
     var list = JSON.parse(data); 
     for(var i = 0; i < list.length; i++){ 
    $('#tanggal').val((list[i]['s_tanggal'])); 
    $('#nama').val((list[i]['silo'])); 
    var tr = "<tr>"; 
    tr += "<td>"+list[i]['no']+"</td>"; 
    tr += "<td>"+list[i]['sensor1']+"</td>"; 
    tr += "<td>"+list[i]['sensor2']+"</td>"; 
    tr += "<td>"+list[i]['sensor3']+"</td>"; 
    tr += "<td>"+list[i]['sensor4']+"</td>"; 
    tr += "<td>"+list[i]['sensor5']+"</td>"; 
    tr += "<td>"+list[i]['sensor6']+"</td>"; 
    tr += "<td>"+list[i]['sensor7']+"</td>"; 
    tr += "<td>"+list[i]['sensor8']+"</td>"; 
    tr += "<td>"+list[i]['sensor9']+"</td>"; 
    tr += "<td>"+list[i]['sensor10']+"</td>"; 
    tr += "<td>"+list[i]['sensor11']+"</td>"; 
    tr += "<td>"+list[i]['sensor12']+"</td>"; 
    tr += "<td>"+list[i]['ambien']+"</td>"; 
    tr += "<td>"+list[i]['average']+"</td>"; 
    tr += "<td>"+list[i]['deffiasi']+"</td>"; 
    tr += "<td>"+list[i]['status']+"</td>"; 

    tr += "</tr>"; 

    $("#table_s tbody").append(tr); 
    } 
    return false; 
} 
}); 
        } 

    clearInput(); 
}); 
$("#form_input").submit(function() { 
    return false; 
}); 
function clearInput() { 
    $("#form_input :input").each(function() { 
     $('#s1').val(''); 
     $('#s2').val(''); 
     $('#s3').val(''); 
     $('#s4').val(''); 
     $('#s5').val(''); 
     $('#s6').val(''); 
     $('#s7').val(''); 
     $('#s8').val(''); 
     $('#s9').val(''); 
     $('#s10').val(''); 
     $('#s11').val(''); 
     $('#s12').val(''); 
     $('#amb').val('35.0'); 
     $('#avr').val(''); 
     $('#deff').val(''); 
     $('#sts').val(''); 
    }); 


} 

<?php 
/* Attempt MySQL server connection. Assuming you are running MySQL 
server with default setting (user 'root' with no password) */ 
$link = mysqli_connect("localhost", "root", "", "silo"); 


// Check connection 
if($link === false){ 
    die("ERROR: Could not connect. " . mysqli_connect_error()); 
} 

    $no= $_POST['no']; 
    $tanggal= $_POST['datepicker']; 
    $silo= $_POST['silo']; 
    $sensor1= $_POST['s1']; 
    $sensor2= $_POST['s2']; 
    $sensor3= $_POST['s3']; 
    $sensor4= $_POST['s4']; 
    $sensor5= $_POST['s5']; 
    $sensor6= $_POST['s6']; 
    $sensor7= $_POST['s7']; 
    $sensor8= $_POST['s8']; 
    $sensor9= $_POST['s9']; 
    $sensor10= $_POST['s10']; 
    $sensor11= $_POST['s11']; 
    $sensor12= $_POST['s12']; 
    $amb= $_POST['amb']; 
    $avr= $_POST['avr']; 
    $deff= $_POST['deff']; 
    $sts= $_POST['sts']; 
// Attempt insert query execution 
$sql = "INSERT INTO temp1 (no, tanggal, silo, sensor1, sensor2, sensor3, sensor4, sensor5, sensor6, sensor7, sensor8, sensor9, sensor10, sensor11, sensor12, ambien, average, deffiasi, status) VALUES ('$no','$tanggal','$silo','$sensor1','$sensor2','$sensor3','$sensor4','$sensor5','$sensor6','$sensor7','$sensor8','$sensor9','$sensor10','$sensor11','$sensor12','$amb','$avr','$deff','$sts')"; 
if(mysqli_query($link, $sql)){ 
    echo "Records inserted successfully."; 
} else{ 
    echo "Records inserted failed "; 
} 
// Data for Titik1 
$query = mysqli_query($link,"SELECT * FROM temp1"); 
$rows = array(); 

while($tmp= mysqli_fetch_array($query)) { 
    $rows[] = $tmp; 
} 

echo json_encode($rows); 
mysqli_close($link); 
?> 

Answer 1

我覺得你有你的味精存儲在您的排陣這樣

<?php 

$link = mysqli_connect("localhost", "root", "", "silo"); 

$rows = array(); 

if($link === false){ 
die("ERROR: Could not connect. " . mysqli_connect_error()); 
} 

$no= $_POST['no']; 
$tanggal= $_POST['datepicker']; 
$silo= $_POST['silo']; 
$sensor1= $_POST['s1']; 
$sensor2= $_POST['s2']; 
$sensor3= $_POST['s3']; 
$sensor4= $_POST['s4']; 
$sensor5= $_POST['s5']; 
$sensor6= $_POST['s6']; 
$sensor7= $_POST['s7']; 
$sensor8= $_POST['s8']; 
$sensor9= $_POST['s9']; 
$sensor10= $_POST['s10']; 
$sensor11= $_POST['s11']; 
$sensor12= $_POST['s12']; 
$amb= $_POST['amb']; 
$avr= $_POST['avr']; 
$deff= $_POST['deff']; 
$sts= $_POST['sts']; 

$sql = "INSERT INTO temp1 (no, tanggal, silo, sensor1, sensor2, sensor3, sensor4, sensor5, sensor6, sensor7, sensor8, sensor9, sensor10, sensor11, sensor12, ambien, average, deffiasi, status) VALUES ('$no','$tanggal','$silo','$sensor1','$sensor2','$sensor3','$sensor4','$sensor5','$sensor6','$sensor7','$sensor8','$sensor9','$sensor10','$sensor11','$sensor12','$amb','$avr','$deff','$sts')"; 
if(mysqli_query($link, $sql)){ 
$rows['msg']="Records inserted successfully."; 
} else{ 
$rows['msg']="Records inserted failed "; 
} 

$query = mysqli_query($link,"SELECT * FROM temp1"); 


while($tmp= mysqli_fetch_array($query)) { 
$rows[] = $tmp; 
} 

echo json_encode($rows); 
mysqli_close($link); 
?>