PHP mysqli插入命令

Question

我試圖通過數組插入多個值到MySQL中，但它不工作或傳遞錯誤消息，所以我不知道我要去哪裏出錯。任何幫助，將不勝感激。

這裏就是我所說的功能

$testArrayList = array(); 
      $testArrayList[] = 'Account_idAccount'; 
      $testArrayList[] = 'firstName'; 
      $testArrayList[] = 'lastName'; 
      $testArrayValues = array(); 
      $testArrayValues[] = $idAccount; 
      $testArrayValues[] = $firstName; 
      $testArrayValues[] = $lastName; 
      $dbManager->insertValues("User", $testArrayList, $testArrayValues); 

現在，這裏是功能可按被稱爲insertValues。

 public function insertValues($table, $cols, $values) { 
    foreach ($cols as $col) 
     $colString .= $col.','; 
    foreach ($values as $value) 
    { 
     $valueAmount .= '?,'; 
     $valueType .= 's'; 
     $valueParam .= $value.","; 
    } 
    $colString = substr($colString, 0, -1); 
    $valueAmount = substr($valueAmount, 0, -1); 
    $valueParam = substr($valueParam, 0, -1); 

    $mysqli = new mysqli(DBHOST, DBUSER, DBPASSWORD, DBDATABASE); 
    $sql = "INSERT INTO $table ($colString) VALUES($valueAmount)"; 
    /* Prepared statement, stage 1: prepare */ 
    if (!($stmt = $mysqli->prepare($sql))) { 
     echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error; 
    } 
    print_r($valueParam); 
    /* Prepared statement, stage 2: bind and execute */ 
    if (!$stmt->bind_param("$valueType", $valueParam)) { 
     echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 

    if (!$stmt->execute()) { 
     echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error; 
    } 
    /* explicit close recommended */ 
    $stmt->close(); 
    $mysqli->close(); 
} 

Answer 1

有一堆錯誤出現的，這裏有一個重寫的版本，你的功能應該工作：

public function insertValues($table, array $cols, array $values) { 

    $mysqli = new mysqli(DBHOST, DBUSER, DBPASSWORD, DBDATABASE); 

    $colString = implode(', ', $cols); // x, x, x 
    $valString = implode(', ', array_fill(0, count($values), '?')); // ?, ?, ? 

    $sql = "INSERT INTO $table ($colString) VALUES($valString)"; 
    if (!$stmt = $mysqli->prepare($sql)) 
     echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error; 

    foreach ($values as $v) 
     if (!$stmt->bind_param('s', $v)) 
      echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 

    if (!$stmt->execute()) 
     echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error; 

    $stmt->close(); 
    $mysqli->close(); 

} 

你應該還有一次是在構造函數，而不是每種方法初始化的mysqli連接：

：

public function __construct() { 
    $this->mysqli = new mysqli(DBHOST, DBUSER, DBPASSWORD, DBDATABASE); 
} 

public function __destruct() { 
    $this->mysqli->close(); 
} 

而且它，你創建一個合適的函數來處理這些錯誤，如很好

public function showError($message, object $obj) { 
    echo "$message: (" . $obj->errno . ") " . $obj->error; 
} 

導致你這個清潔版本功能：

public function insertValues($table, $cols, $values) { 

    ... 

    if (!$stmt = $mysqli->prepare($sql)) 
     $this->showError("Prepare failed", $mysqli); 

    foreach ($values as $v) 
     if (!$stmt->bind_param('s', $v)) 
      $this->showError("Binding parameters failed", $stmt); 

    if (!$stmt->execute()) 
     $this->showError("Execute failed", $stmt); 

    ... 

} 

Answer 2

我已經重寫功能讓您可以清楚瑟爲什麼你使用bind_param（）是錯誤的。

這個版本只是一個例子，它僅適用於2列！

function insertValues($table, array $cols, array $values) { 

     $mysqli = new mysqli('localhost', 'petr', null,'test'); 

     $colString = implode(', ', $cols); // x, x, x 
     $valString = implode(', ', array_fill(0, count($values), '?')); // ?, ?, ? 

     $sql = "INSERT INTO $table ($colString) VALUES($valString)"; 
     if (!$stmt = $mysqli->prepare($sql)) 
      echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error; 

     list($a,$b) = $values; 
     // params $a and $b must exists during $stmt execution, therefore you can't use foreach with temproray variable 
     if (!$stmt->bind_param('ss', $a, $b)) 
       echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 

     if (!$stmt->execute()) 
      echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error; 

     $stmt->close(); 
     $mysqli->close(); 

    } 

這工作：

insertValues('test',array('firstName','lastName'),array('Jan Amos','Komensky'));