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[英]Concat two columns from mysql and display distinct values in dropdown list
[英]concat two columns
如何合并两列? 我需要FIRSTNAME和LASTNAME等于$ Employee_Name我尝试了不同的组合,但仍然无法使用。 我也知道这有一个巨大的SQL注入问题。 我尚不知道如何解决我要查找的问题。
<?php
header('Content-Type: application/json');$db_conx = mysqli_connect("localhost", "root", "systems399","employees_db");
$Employee_Name= $_POST["Employee_Name"];
$sql="SELECT * FROM names WHERE FIRSTNAME='$Employee_Name' ";
$query= mysqli_query($db_conx, $sql);
$row = mysqli_fetch_array($query, MYSQLI_ASSOC);
$rc= $row["EMPLOYEE_NUMBER"];
echo json_encode ($rc);
?>
我以这种方式尝试过,但没有用。
<?php
header('Content-Type: application/json');
$db_conx = mysqli_connect("localhost", "root", "systems399", "employees_db");
$Employee_Name= $_POST["Employee_Name"];
$sql="SELECT * FROM names WHERE FIRSTNAME='$Employee_Name' AND LASTNAME='$Employee_Name'";
$query= mysqli_query($db_conx, $sql);
$row = mysqli_fetch_array($query, MYSQLI_ASSOC);
$rc= $row["EMPLOYEE_NUMBER"];
echo json_encode ($rc);
?>
这是我的Javascript代码
$(document).ready(function() {
$("#Employee_Name").change(function() {
var Employee_Name = $(this).val();
if (Employee_Name != '') {
$.ajax({
type: "post",
url: "insert.php",
data: "Employee_Name=" + Employee_Name,
datatype: "json",
success: function(data) {
$("#Employee_Number").val(data);
$('#Employee_Number').css("background-color", "#B3CBD6")
$('#Employee_Number').animate({
backgroundColor: "#ffffff"
});
},
error: function(response) {
alert("error scripting")
}
});
} else {
$("#Employee_Number").val("");
}
});
});
有两种可能性:
1。
$sql = "SELECT CONCAT(`FIRSTNAME`, ' ', `LASTNAME`) AS `EmployeeName`, * FROM `names` HAVING `EmployeeName` = '".$Employee_name."'";
2。
$sql = "SELECT * FROM `names` WHERE CONCAT(`FIRSTNAME`, ' ', `LASTNAME`) = '".$Employee_name."'";
这是一个例子-
$sql = "SELECT CONCAT(`FIRSTNAME`, ' ', `LASTNAME`) AS `EmployeeName`, * FROM `names` WHERE `FIRSTNAME` = '" . $First_Name . "' AND `LASTNAME` = '" . $Last_Name' . "' ";
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