php - 数据库中所有列中仅插入一个值

Question

require_once 'core.php' ;

try {
 $db = new PDO("mysql:host=$host;dbname=$dbname",$user,$password) ;
 $db -> setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);

 $a = array(
    'name' => 'rocky',
    'password' => '12345' ,
    'age' => '22'
    ); 
 $c = implode(",",array_keys($a) ) ;
 $f = ":".implode(" , :",array_keys($a));

  $db->beginTransaction();
  $query = $db->prepare("INSERT INTO try ($c) VALUES ($f)");
  foreach ($a as $key => $value) {
        $query->bindParam(":".$key,$value,PDO::PARAM_STR) ;
  } 
  $query->execute() ;
  $db->commit() ;


} catch(PDOException $e){
die($e->getMessage()) ;
} 

在数据库中，只有一个年龄值插入到名称、密码、年龄列中，与 22 ,22 ,22 等值相同

Answer 1

问题出在 bindParam 函数中:

$db->beginTransaction();
$query = $db->prepare("INSERT INTO try ($c) VALUES ($f)");
foreach ($a as $key => $value) {
    $query->bindParam(":".$key,$value,PDO::PARAM_STR) ;
} 
$query->execute() ;
$db->commit() ;

阅读bindParam手册:

Unlike PDOStatement::bindValue(), the variable is bound as a reference and will only be evaluated at the time that PDOStatement::execute() is called.

这意味着当您执行时 - 所有绑定(bind)变量都设置为 最后一个$value值是数组的最后一个值，是22。

因此，使用bindValue:

$db->beginTransaction();
$query = $db->prepare("INSERT INTO try ($c) VALUES ($f)");
foreach ($a as $key => $value) {
    $query->bindValue(":".$key, $value, PDO::PARAM_STR) ;
} 
$query->execute() ;
$db->commit() ;