[英]Value is not getting fetched from database using mysqli bind_result
我试图使用mysqli获取在我的数据库表中找到的int
。 然后我使用bind_result()
绑定结果。 但是,当我尝试使用该值时,我只需得到0.我该如何处理?
我的代码看起来像这样:
$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\">";
}
现在,它不会在输入框内显示10。 这意味着它不执行if语句。 但是,为什么我没有从$db_salary_amount
获得任何结果? 我知道我可能不必将变量从$db_salary_amount
为$db_salary_amount
。 我只是想找出问题所在。
你在bind_result()
之后缺少一个->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;
来自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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