你好,我的PHP代码中有一个烦人的错误。我问了我的老师,他的意思是代码应该工作,他不知道为什么会出现这个错误。我希望您能找到错误:
在这段代码中,我试图成为json格式:
<?php
$von = $_GET['datumvon'];
$bis = $_GET['datumbis'];
$frei =0;
$DB_server = "localhost";
$DB_name = "dbterminkalenderha";
$DB_user = "root";
$DB_password = "root";
$server = mysqli_connect($DB_server, $DB_user, $DB_password);
$connection = mysqli_select_db($server, $DB_name);
$myquery = $server->prepare("SELECT T_Datum,T_Frei FROM t_terminslots WHERE (T_Datum Between ? AND ?) AND (T_Frei = ?); ");
$myquery->bind_param("ssi",$von,$bis,$frei);
$myquery ->execute();
//$query = mysqli_($myquery);
if ( ! $myquery ) {
echo mysqli_error();
die;
}
$data = array();
for ($x = 0; $x < mysql_num_rows($myquery); $x++) {
$data[] = mysqli_fetch_assoc($myquery);
}
echo json_encode($data);
mysqli_close($server);
?>
错误:警告:mysql_num_rows()要求参数1是资源,在
您需要使用mysqli函数。
for ($x = 0; $x < mysqli_num_rows($myquery); $x++) {
$data[] = mysqli_fetch_assoc($myquery);
}