我有此HTML代码的形式,我想检查用户输入的值在数据库中是否已经存在。
<tr height="50">
<td valign="top"><label for="CINumber"><span style="color:red">*</span>CI Name:</label></td>
<td><input type="text" id="CINumber" name="CINumber"></td>
</tr>
我有这个``JavaScript函数用于汇总表单值:
function submit(){
var cNumber = document.getElementById("CINumber").value;
//many form value within this function but I showed this value only cause this value need to check when click submit button.
$.ajax({
url: 'PHPMethodCalls_AL.php',
type: 'post',
data: {'action':'AddNewCI', 'cNumber' : cNumber },
success: function(data,status) {
//alert(data);
if(status == "success"){
alert("CI Name can be assigned");
alert("Successfully Register New CI Record.");
} else {
alert("CI Name is already present!");
return ;
}
},
error: function(xhr, desc, err) {
console.log(xhr);
console.log("Details: " + desc + "\nError:" + err);
}
});
}
然后在“ PHPMethodCalls_AL.php” URL中调用PHPMethod调用。
if($_POST['action'] == "AddNewCI"){
$cNumber = $_POST['cNumber'];
$newCI = AddNewCI($cNumber);
echo json_encode($newCI );
}
之后,我调用与数据库连接的
PHP
函数。function AddNewCI($cNumber){
global $DB;
if(!empty($cNumber)){
$namequery="SELECT Name from Item where Name='$cNumber'";
$namecheck=mysql_query($namequery);
$row=mysql_num_rows($namecheck);
if($row==0){
echo "success";
$strSql1 ="INSERT INTO Item (Name)VALUES('$cNumber')";
$res1 = $DB->Query($strSql1, false, $err_mess.__LINE__);
}
else {
echo "!success";
}
}
}
单击提交后,我只得到了(状态==“成功”),然后弹出2个警告框,即使数据库中已经存在数据。但是,重复值未保存在数据库中,因为我为此字段设置了UNIQUE。如果数据存在,请建议我如何获取正确的警报框;如果不存在,则显示“ Alredy存在”;如果不显示“显示成功警报框” ...
最佳答案
更改以下代码...
url: 'PHPMethodCalls_AL.php',
type: 'post',
data: {'action':'AddNewCI', 'cNumber' : cNumber },
success: function(data) {
if(data == "success"){
alert("CI Name can be assigned");
alert("Successfully Register New CI Record.");
}
无需使用json_encode,您可以根据需要直接回显任何值。
if($_POST['action'] == "AddNewCI"){
$cNumber = $_POST['cNumber'];
$newCI = AddNewCI($cNumber);
echo $newCI;
}
更改函数的返回值...
function AddNewCI($cNumber){
global $DB;
if(!empty($cNumber)){
$namequery="SELECT Name from Item where Name='$cNumber'";
$namecheck=mysql_query($namequery);
$row=mysql_num_rows($namecheck);
if($row==0){
$strSql1 ="INSERT INTO Item (Name)VALUES('$cNumber')";
$res1 = $DB->Query($strSql1, false, $err_mess.__LINE__);
return "success";
}
else
{
return "error";
}
}
}