Php 将$item=mysql\u fetch\u assoc($stmt)更改为prepared语句样式
这段代码可以工作,但我正在尝试如何更改$rose=mysql\u fetch\u assoc($stmt);“准备好的语句样式”部分。有人知道吗Php 将$item=mysql\u fetch\u assoc($stmt)更改为prepared语句样式,php,sql,prepared-statement,Php,Sql,Prepared Statement,这段代码可以工作,但我正在尝试如何更改$rose=mysql\u fetch\u assoc($stmt);“准备好的语句样式”部分。有人知道吗 $rose_id = $_GET['rose_id']; //prepare the statement $stmt = $conn2->prepare("SELECT * FROM rosename LEFT JOIN rosevariety ON (rosename.variety_name = roseva
$rose_id = $_GET['rose_id'];
//prepare the statement
$stmt = $conn2->prepare("SELECT * FROM rosename
LEFT JOIN rosevariety ON (rosename.variety_name = rosevariety.variety_name)
WHERE rose_id = ?");
//bind the parameters
$stmt->bind_param("i", $rose_id);
//$sql = mysql_query($query, $conn);
$stmt->execute();
//was there a good response?
if ($stmt) {
$rose = mysql_fetch_assoc($stmt);
//echo out rose information
echo "<h1>".$rose['latin_name']."</h1>";
echo "<h2>".$rose['common_name']."</h2>";
$rose\u id=$\u GET['rose\u id'];
//准备声明
$stmt=$conn2->prepare(“从rosename中选择*
左键连接rosevariety ON(rosename.variety\u name=rosevariety.variety\u name)
其中rose_id=?”;
//绑定参数
$stmt->bind_参数(“i”,$rose_id);
//$sql=mysql\u查询($query,$conn);
$stmt->execute();
//反应好吗?
如果($stmt){
$rose=mysql\u fetch\u assoc($stmt);
//呼出玫瑰的信息
echo“$rose[“拉丁名]”;
echo“$rose['common_name']”;
如果使用PDO:
$rose = $stmt->fetch(PDO::FETCH_ASSOC);
如果使用mysqli:
$result = $stmt->get_result();
$rose = $result->fetch_assoc();
如果使用PDO,则(PDO::FETCH_ASSOC)将是答案。 事实上,我认为要正确使用它,你需要
while(($rose = $stmt->fetch()) !== false){
//$rose = current row;
}
这看起来像是
mysqli
而不是PDO
是它的mysqli。我如何为mysqli做到这一点?
while(($rose = $stmt->fetch()) !== false){
//$rose = current row;
}