[英]Value is not getting fetched from database using mysqli bind_result
我試圖使用mysqli獲取在我的數據庫表中找到的int
。 然后我使用bind_result()
綁定結果。 但是,當我嘗試使用該值時,我只需得到0.我該如何處理?
我的代碼看起來像這樣:
$sql = <<<EOF
SELECT
project_salary_amount
FROM projects_set_salary
WHERE project_id = ? ORDER BY project_salary_id DESC LIMIT 1
EOF;
$stmt = $mysqli->prepare($sql) or die ("Feil i database<br>" . $sql . "<br><b>Feilmelding:</b> " . $mysqli->error);
$stmt->bind_param("i", $project_id);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($dbb_salary_amount);
$db_salary_amount = $dbb_salary_amount;
$num_salary_results = $stmt->num_rows;
$stmt->free_result();
$stmt->close();
if($num_salary_results == 0){
$sql = <<<EOF
INSERT INTO
projects_set_salary (project_id)
VALUES ($project_id)
EOF;
$stmt = $mysqli->prepare($sql) or die ("Feil i database<br>" . $sql . "<br><b>Feilmelding:</b> " . $mysqli->error);
$stmt->execute();
$stmt->close();
$db_salary_amount = 10;
$html_set_salary = "";
$html_set_salary .= "<form id=\"form_send_salary\"method=\"post\" action=\"create_set_salary.php\">\n";
$html_set_salary .= "<input type=\"number\" name=\"set_salary\" value=\"$db_salary_amount\">";
}
else{
$html_set_salary = "";
$html_set_salary .= "<form id=\"form_send_salary\"method=\"post\" action=\"create_set_salary.php\">\n";
$html_set_salary .= "<input type=\"number\" name=\"set_salary\" value=\"$db_salary_amount\">";
}
現在,它不會在輸入框內顯示10。 這意味着它不執行if語句。 但是,為什么我沒有從$db_salary_amount
獲得任何結果? 我知道我可能不必將變量從$db_salary_amount
為$db_salary_amount
。 我只是想找出問題所在。
你在bind_result()
之后缺少一個->fetch()
bind_result()
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($dbb_salary_amount);
$stmt->fetch(); // <--- missing this ->fetch()
$db_salary_amount = $dbb_salary_amount;
來自docs for ->bind_result()
- > When mysqli_stmt_fetch() is called to fetch data, the MySQL client/server protocol places the data for the bound columns into the specified variables var1, ....
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