我只是一個初學者,學習如何編寫代碼。我正在關注一個cbt塊,我已經在一個操作上將插入的值發佈到表單上,並將其傳遞到操作頁面並插入到數據庫中。我得到這個消息時,我嘗試提交按鈕Post函數在php中不起作用
0Problems與查詢:
我的DB連接器
<?php
$host = "localhost";
$user = "root";
$password = "";
$db = "linuxcbtcontacts";
# Step 1 instantiate DB object
$conn = mysqli_connect($host,$user,"",$db) or die("Problems connecting:" . mysql_error());
#if ($conn) { echo "True"; }
#$dbselect = mysqli_select_db("linuxcbtcontacts", $conn) or die("Error selecting DB:" . mysql_error());
# Step 2 - select DB
#$dbselect = mysqli_select_db($conn, "linuxcbtcontacts") or die("Error selecting DB:" . mysql_error());
$dbselect = mysqli_select_db($conn,$db) or die("Error selecting DB:" . mysql_error());
?>
我的表單操作頁面
<?php
require 'dbconnect.php';
$unique_check = "select name from contacts where name = 'name'";
$results_unique_check = mysqli_query($conn,$unique_check);
$rowcount = mysqli_num_rows($results_unique_check);
echo $rowcount;
if ($rowcount > 0) {
echo "Please use in different name", "<br>";
echo "<a href=query2.php>go back</a>";
exit(); }
else {
$query1 = "INSERT INTO contacts(name,email,age,yearborn,ratesite,industry,buyourproducts,software,hardware) VALUES('$_POST[name]','$_POST[email]','$_POST[age]','$_POST[yearborn]','$_POST[ratesite]','$_POST[industry]','$_POST[buyourproducts],'$_POST[software]','$_POST[hardware]')";
# Step 3 - invoke query
$results = mysqli_query($conn,$query1) or die("Problems with query:" . mysql_error());
$rowcount = mysqli_affected_rows($conn);
echo "Total Inserted Records:\t", $rowcount, "<br>";
}
?>
(打從表單提交後)
從cbt我應該得到這個消息提交按鈕被點擊後
0Total Inserted Records:1
已經調整了所有準備什麼我在互聯網上看到了但我仍然結束了「與查詢0Problems:」消息
不能混搭'mysql_ *'和'mysqli_ *功能。 –
對所有POST值更改'VALUES('$ _ POST [name]',..'到'VALUES('「。$ _ POST ['name']。'',') –
@NanaPartykar你試圖說的是什麼區別將使? –