問題描述
我正在使用 mysqli prepare 語句來查詢具有多個約束的數據庫.我已經在我的測試文件中運行了代碼，它運行良好.但是，當我將代碼移至實時文件時，它會引發以下錯誤:
I am using a mysqli prepare statement to query my db with multiple constraints. I have ran the code in a test file of mine and it works perfectly fine. However, when I move the code over to my live file it throws the error below:
PHP 警告:mysqli_stmt::bind_result():綁定變量的數量與準備好的語句中的字段數不匹配C:wampwwwfirecomfirecom.php 80 行
PHP Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in C:wampwwwfirecomfirecom.php on line 80
PHP 注意:未定義變量:導致C:wampwwwfirecomfirecom.php 第 89 行
PHP Notice: Undefined variable: results in C:wampwwwfirecomfirecom.php on line 89
兩個參數都設置正確，但有些東西把它扔掉了.
Both parameters are being set correctly but something is throwing it off.
代碼:
$query = $mysqli->prepare("SELECT * FROM calls WHERE wcccanumber = ? && county = ?");
$query->bind_param("ss", $wcccanumber, $county);
$query->execute();
$meta = $query->result_metadata();
while ($field = $meta->fetch_field()) {
$parameters[] = &$row[$field->name];
}
call_user_func_array(array($query, 'bind_result'), $parameters);
while ($query->fetch()) {
foreach($row as $key => $val) {
$x[$key] = $val;
}
$results[] = $x;
}
print_r($results['0']);
$query var_dump:
$query var_dump:
object(mysqli_stmt)#27 (10) { ["affected_rows"]=> int(-1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(2) ["field_count"]=> int(13) ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["sqlstate"]=> string(5) "00000" ["id"]=> int(1) }
推薦答案
為什么要用 mysqli 折磨自己?
在 PDO 中，您不需要這些可怕的代碼，只需要一行代碼即可獲得結果
Why torture yourself with mysqli?
In PDO you will need none of these horrendous codes, but only one line to get the results
$query = $pdo->prepare("SELECT * FROM calls WHERE wcccanumber = ? && county = ?");
$query->execute(array($wcccanumber, $county));
$results = $query->fetchAll();
print_r($results[0]);
這篇關于PHP mysqli準備語句不起作用的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！