無法連接到MySQL的PHP​​

Question

新的PHP和我連接形式屬性到PHP連接到一個godaddy MySQL。每一次嘗試都以沒有錯誤信息的黑屏結束。跳出來有沒有語法錯誤？我崇高的文本不會註冊PHP的語法，但這又是另一個問題。我可能需要調用godaddy支持？該密碼已被刪除的隱私。根據PHP代碼標準每行結束後，

<?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

請連接數據庫是這樣的...

$connection = mysqli_connect(DB_SERVER,DB_USER,DB_PASS); 
if (!$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 並插入分號（）。

Answer 2

此代碼有很多問題，如mysqli_select_db問題所述。 $_POST['BIx']也會導致錯誤，因爲每個語句後都沒有分號。你錯過了'（'線if ($conn->query<$sql) === TRUE) {更不用說那條線不會工作，因爲你邏輯上比較資源類型（我認爲）與字符串。

你也永遠不會執行插入。聲明一切圍繞我認真地想你應該練PHP編碼更多一些，並就如何正確使用mysqli的閱讀起來：see here

問候

編輯：您還可以關閉PHP的標籤在你的腳本結束這通常不是一個好主意，如解釋here

編輯2：也使用IDE，比如Netbeans始終是因爲它可以突出語法錯誤，而不是提了要爲你做一個好主意）

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(); 

?>