我想改变我的mysqli检查存在并插入到pdo ..我的声明如下所示:
<?php
$addSearch = preg_replace('/[^a-zA-Z0-9_ %\[\]\.\(\)%&-]/s', '', $search);
if ($CheckMusic = $mysqli->query("SELECT * FROM search WHERE term='$addSearch'")) {
$CheckRow = mysqli_fetch_array($CheckMusic);
$CheckId = $CheckRow['id'];
$CheckCnt = $CheckMusic->num_rows;
$CheckMusic->close();
} else {
printf("Error: %s\n", $mysqli->error);
}
$Now = strtotime("now");
$addSearch = $mysqli->escape_string($addSearch);
if ($CheckCnt < 1) {
$mysqli->query("INSERT INTO search (term, datetime) VALUES('$addSearch','$Now') ") or die(mysqli_error());
} else {
$mysqli->query("UPDATE search SET datetime='$Now' WHERE id='$CheckId'") or die(mysqli_error());
}
?>
希望有人可以提供样品帮助; q谢谢你的预付款!
答案 0 :(得分:0)
<?php
$addSearch = preg_replace('/[^a-zA-Z0-9_ %\[\]\.\(\)%&-]/s', '', $search);
$CheckMusic = $mysqli->query("SELECT * FROM search WHERE term='$addSearch'");
$Now = strtotime("now");
if($CheckMusic->num_rows == 0){
$mysqli->query("INSERT INTO search (term, datetime) VALUES('$addSearch','$Now') ") or die(mysqli_error());
}else{
$CheckRow = mysqli_fetch_array($CheckMusic);
$CheckId = $CheckRow['id'];
$addSearch = $mysqli->escape_string($addSearch);
$mysqli->query("UPDATE search SET datetime='$Now' WHERE id='$CheckId'") or die(mysqli_error());
}
我希望它能奏效。