Question

新的php并将表单属性连接到php以连接到godaddy mysql。每次尝试都以空白屏幕结束，没有错误消息。跳出来有没有语法错误？我的崇高文本不会注册php语法，但这是另一个问题。我可能需要打电话给godaddy支持吗？密码已被删除以保护隐私。

<?php

$servername = "localhost";
$dbusername = "jaysenhenderson";
$dbpassword = "xxxxx";
$dbname = "EOTDSurvey";


$con = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
mysql_select_db('EOTDSurvey', $con)

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo("Connected successfully");


$_POST['BI1']
$_POST['BI2']
$_POST['BI3']
$_POST['BI4']
$_POST['BI5']
$_POST['BI6']
$_POST['BI7']
$_POST['BI8']
$_POST['BI9']
$_POST['BI10']
$_POST['BI11']
$_POST['BI12']
$_POST['BI13']
$_POST['BI14']
$_POST['BI15']

$sql = "INSERT INTO Survey1(BI1)"
$sql = "INSERT INTO Survey1(BI2)"
$sql = "INSERT INTO Survey1(BI3)"
$sql = "INSERT INTO Survey1(BI4)"
$sql = "INSERT INTO Survey1(BI5)"
$sql = "INSERT INTO Survey1(BI6)"
$sql = "INSERT INTO Survey1(BI7)"
$sql = "INSERT INTO Survey1(BI8)"
$sql = "INSERT INTO Survey1(BI9)"
$sql = "INSERT INTO Survey1(BI10)"
$sql = "INSERT INTO Survey1(BI11)"
$sql = "INSERT INTO Survey1(BI12)"
$sql = "INSERT INTO Survey1(BI13)"
$sql = "INSERT INTO Survey1(BI14)"
$sql = "INSERT INTO Survey1(BI15)"

if ($conn->query<$sql) === TRUE) {
    echo "IT FUCKING WORKS.";
}
else{
    echo "didnt workkkkkk";
}
$conn->close();

 ?>

Answer 1

请像这样连接数据库...

$connection = mysqli_connect(DB_SERVER,DB_USER,DB_PASS);
if (!$connection) {
die("Database connection failed: " . mysqli_error());
}

// 2. Select a database to use 
$db_select = mysqli_select_db($connection, DB_NAME);
if (!$db_select) {
die("Database selection failed: " . mysqli_error());
}

并使用mysqli_select_db而不是mysql_select_db 并根据php代码标准在每行结束后插入分号（;）。

Answer 2

此代码存在很多问题，如mysqli_select_db问题所述。 $_POST['BIx']也会导致错误，因为每个语句后都没有分号。你错过了一个'（'就行if ($conn->query<$sql) === TRUE) {，更不用说那条线无论如何都行不通，因为你在逻辑上将资源类型（我认为）与字符串进行比较。

您也永远不会执行insert语句。我认真地认为你应该更多地练习PHP编码并阅读如何正确使用mysqli：see here。

此致

编辑：你的脚本末尾还有一个结束的PHP标签，这通常不是一个好主意，如here所述

编辑2：同样使用诸如Netbeans之类的IDE总是一个好主意，因为它可以突出显示语法错误，而不是要求SO为您执行此操作;）

Answer 3

<?php

$servername = "localhost";
$dbusername = "jaysenhenderson";
$dbpassword = "xxxxx";
$dbname = "EOTDSurvey";


$con = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
mysqli_select_db('EOTDSurvey', $con);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo("Connected successfully");


############# Function For Insert ##############


function insert($tableName='',$data=array())
  {
    $query = "INSERT INTO `$tableName` SET";
    $subQuery = '';
    foreach ($data as $columnName => $colValue) {
      $subQuery .= " `$columnName`='$colValue',";
    }
    $subQuery = rtrim($subQuery,', ');
    $query .= $subQuery;
    pr($query);
    mysqli_query($con,$query) or die(mysqli_error());
    return mysqli_insert_id();
  }//end insert

#########################################  
if(isset($_POST['submit'])){
    unset($_POST['submit']);
    //print_r($_POST);
    $result=insert('Survey1',$_POST);

if($result){

 echo '<script>window.alert("Success!");</script>';
 echo "<script>window.location.href = 'yourpage.php'</script>";
}   
}

$conn->close();

 ?>

无法将php连接到mysql

3 个答案: