我如何从PHP中的数据库中获取数据?

时间:2016-05-13 06:24:35

标签: php

这里公司名称的总行数是100000,但得到的错误就像 服务器无法处理请求。

  <?php
  $servername="localhost";
  $username="root";
  $password="";
  $dbname="ABC";
  $conn = mysqli_connect($servername, $username, $password, $dbname );
  if($conn){  
  $result = mysqli_query($conn, " SELECT COMPANY_NAME from mytask "));
$emparray = array();    
  while($row = mysqli_fetch_assoc($result))    
   {
    $emparray[] = $row;        
   }  
   var_dump($emparray);
   exit; 
}
else{
 echo "Connection Not Successful"; 
}
?>

2 个答案:

答案 0 :(得分:-1)

你有一个拼写错误,一个额外的括号。更改以下行:

$result = mysqli_query($conn, " SELECT COMPANY_NAME from mytask "));

到这一行:

$result = mysqli_query($conn, " SELECT COMPANY_NAME from mytask ");

<强>编辑: 另外,请更改以下代码:

  while($row = mysqli_fetch_assoc($result))    
   {
    $emparray[] = $row;        
   }  
   var_dump($emparray);

进入此代码:

  while($row = mysqli_fetch_assoc($result))    
   {
      foreach ($row as $value)
      {
         echo "--".$value;
      }
      echo "<br>";     
   }  

让我知道它现在是否有效。

答案 1 :(得分:-1)

你可以尝试:

$mysqli = new mysqli($servername,$username,$password,$dbname);
$stmt = $mysqli->prepare("SELECT COMPANY_NAME FROM mytask");
$sql = stmt->execute();
while ($row = $sql->fetch_object()) {
    $emparray[] = $row->COMPANY_NAME;
}
var_dump($emparray);
exit;