[英]Updating database using ajax and php
我正在嘗試將我的數據庫tinyint更新為1(選中該復選框時),將其更新為0(如果取消選中該復選框)。 我的Javascript / Ajax代碼是:
<script>
function emailNextWithAddress(chk,address) {
var nextEmail, inside_where;
if(chk.checked === true){
$.ajax({
url: "marca_enviado.php",
type: "get",
data: address,
success: function(){
nextEmail = document.createElement('input');
nextEmail.id = address;
nextEmail.value = address;
nextEmail.type = 'text';
nextEmail.name = 'email[]';
nextEmail.className = 'insemail';
nextEmail.style.display = 'inline-block';
inside_where = document.getElementById('addEmail');
inside_where.appendChild(nextEmail);
},
error: function(){
console.log("Erro!");
}
});
} else {
$.ajax({
url: "desmarca_enviado.php",
type: "get",
data: address,
success: function(){
inside_where = document.getElementById(address);
inside_where.parentNode.removeChild(inside_where);
},
error: function(){
console.log("Erro!");
}
});
}
return false;
}
</script>
試圖更新我的數據庫tinyint的php(marca_enviado.php)代碼是:
<?php
include_once 'dbh.php';
$address = $_GET['address'];
$updateEnviado = "UPDATE escolas SET Enviado = 1 WHERE id= 'address' ";
$result = mysqli_query($conn , $updateEnviado);
?>
<?php
include_once 'dbh.php';
$address = $_GET['address'];
//-- escape the input data. But be sure to add some validations before this line.
$address = mysqli_real_escape_string($conn, $address);
//-- use "$address" variable there, assuming that variable contains the id
$updateEnviado = "UPDATE escolas SET Enviado = 1 WHERE id= '$address' ";
$result = mysqli_query($conn , $updateEnviado);
?>
希望能幫助到你!
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