Question

让我说我有和像这样的数组

$array= Array('id'=>'3', 'name'=>'NAME', 'age'=>'12');

此数组中的键是表中列的名称，值是我需要更新的列的值。我想根据键和值更新表。我正在使用ADODB 请帮帮我

Answer 1

试试这个：

$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
   $sql .= $key . " = " . $value . ", "; 
}

$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas

当然还有WHERE子句：

$sql .= " WHERE condition = value";

你会得到字符串：

UPDATE table SET id = 3, name = NAME, age = 12 WHERE condition = value

L.E：您可能需要向字符串添加撇号，因此我必须将代码更改为以下内容：

$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
   if(is_numeric($value))
      $sql .= $key . " = " . $value . ", "; 
   else
      $sql .= $key . " = " . "'" . $value . "'" . ", "; 
}

$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas

$sql .= " WHERE condition = value";

将产生这个：

UPDATE table SET id = 3, name = 'NAME', age = 12 WHERE condition = value

L.E 2：如果你想在你的条件中使用id列，代码就变成了这样：

$sql = "UPDATE table SET ";
foreach($array as $key=>$value) {
   if($key == 'id'){
      $sql_condition = " WHERE " . $key . " = " . $value;
      continue;
   }
   if(is_numeric($value))
      $sql .= $key . " = " . $value . ", "; 
   else
      $sql .= $key . " = " . "'" . $value . "'" . ", "; 
}

$sql = trim($sql, ' '); // first trim last space
$sql = trim($sql, ','); // then trim trailing and prefixing commas

$sql .= $sql_condition;

将产生这个结果：

UPDATE table SET name = 'NAME', age = 12 WHERE id = 3

希望这有帮助！：d

Answer 2

foreach ($update_array as $key => $testimonials) {
    $name = mysql_real_escape_string($testimonials->name);
    $content = mysql_real_escape_string($testimonials->content);
    $id = intval($testimonials->id);

    $sql = "UPDATE testimonials SET name='$name', content='$content' WHERE id=$id";
    $result = mysql_query($sql);
    if ($result === FALSE) {
        die(mysql_error());
    }
}

来源：https://stackoverflow.com/a/7884331/3793639

要检查的其他来源。 PHP SQL Update array和Simple UPDATE MySQl table from php array

Answer 3

您可以使用类似的东西来实现这一目标：

foreach($values as $value) {
  if(!key_exists($value, $item)) {
    return false;
  }

    $table->{$value} = $items[$value];
}

Answer 4

假设密钥索引始终为id且adodb可以使用命名占位符，则可以执行以下操作：

$array = Array('id'=>'3', 'name'=>'NAME', 'age'=>'12');
$set = array();
$data = array();
while(list($key,$value)=each($array)) {
    $data[':'.$key] = $value;
    if($key!='id') {
        $set[] = $key . ' = :' . $key;
        // if no placeholders use $set[] = $key . " = '" . database_escape_function($value) . "'";
    }
}
$sql = "UPDATE table SET ".implode($set, ',')." WHERE id=:id";

//$data is now Array(':id'=>'3', ':name'=>'NAME', ':age'=>'12');
//$sql is now  "UPDATE table SET name=:name, age=:age WHERE id=:id";

$stmt = $DB->Prepare($sql);
$stmt = $DB->Execute($stmt, $data);

使用PHP数组进行SQL更新

4 个答案: