无法连接到MySQL的PHP​​

Question

新的PHP和我连接形式属性到PHP连接到一个godaddy MySQL。每一次尝试都以没有错误信息的黑屏结束。跳出来有没有语法错误？我崇高的文本不会注册PHP的语法，但这又是另一个问题。我可能需要调用godaddy支持？该密码已被删除的隐私。根据PHP代码标准每行结束后，

<?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

请连接数据库是这样的...

$connection = mysqli_connect(DB_SERVER,DB_USER,DB_PASS); 
if (!$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()); 
} 

和使用mysqli_select_db代替mysql_select_db 并插入分号（）。

Answer 2

此代码有很多问题，如mysqli_select_db问题所述。 $_POST['BIx']也会导致错误，因为每个语句后都没有分号。你错过了'（'线if ($conn->query<$sql) === TRUE) {更不用说那条线不会工作，因为你逻辑上比较资源类型（我认为）与字符串。

你也永远不会执行插入。声明一切围绕我认真地想你应该练PHP编码更多一些，并就如何正确使用mysqli的阅读起来：see here

问候

编辑：您还可以关闭PHP的标签在你的脚本结束这通常不是一个好主意，如解释here

编辑2：也使用IDE，比如Netbeans始终是因为它可以突出语法错误，而不是提了要为你做一个好主意）

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(); 

?>