使用准备好的语句返回特定行

发布于 2024-08-11 15:25:03 字数 601 浏览 3 评论 0原文

我有一个问题，我准备好的语句似乎只返回返回的行数而不是行的值。下面是我的代码。我确实尝试过谷歌，但它没有告诉我任何东西！如果有人能告诉我我做错了什么以及如何解决它，我将非常感激。谢谢

$query2 = 'SELECT * FROM kids_entry WHERE email = ?';
$stmt2 = $connection->prepare($query2);

// bind the user id as an integer to the first ?
$stmt2->bind_param('s', $email);
$stmt2->execute(); // execute the statement

$stmt2->store_result(); // this call is required for the next operation 

while($row1 = $stmt2->fetch()){
    printf ("%s \n", $entries);
}

编辑

我只是尝试用 while 循环替换 if ，我得到了同样的结果。

编辑2

添加新代码后，它可以工作，但是我如何将其分配给$变量？

原文

I have a problem where my prepared statement appears to only be returning the number of rows returned rather than the value of the row. Below is my code. I did try google for this but it doesn't tell me anything! If someone could tell me what I am doing wrong and how to fix it I would be very appreciative. Thanks

$query2 = 'SELECT * FROM kids_entry WHERE email = ?';
$stmt2 = $connection->prepare($query2);

// bind the user id as an integer to the first ?
$stmt2->bind_param('s', $email);
$stmt2->execute(); // execute the statement

$stmt2->store_result(); // this call is required for the next operation 

while($row1 = $stmt2->fetch()){
    printf ("%s \n", $entries);
}

EDIT

I did just try replacing the if with a while loop and i got the same thing.

EDIT 2

Having added the new code it works but how would i assign this to a $variable?

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一梦等七年七年为一梦 2024-08-18 15:25:03

请参阅 php 文档中的示例：
mysqli_stmt_fetch

/* execute statement */
$stmt->execute();

/* bind result variables */
$stmt->bind_result($name, $code);

/* fetch values */
while ($stmt->fetch()) {
    printf ("%s (%s)\n", $name, $code);
}

/* close statement */
$stmt->close();

See the example in the php documentation:
mysqli_stmt_fetch

/* execute statement */
$stmt->execute();

/* bind result variables */
$stmt->bind_result($name, $code);

/* fetch values */
while ($stmt->fetch()) {
    printf ("%s (%s)\n", $name, $code);
}

/* close statement */
$stmt->close();

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