下面的代码在我的本地服务器上运行时没有任何问题。但是,当我尝试在预期的服务器上运行它时,我的两个查询不起作用 - 它们不是INSERT
,因为它们应该是这样。我标记了两个不支持评论的查询,其余的都有效。目标服务器在PHP 5.6.30-0 + deb8u1上运行。插入查询在服务器上不起作用
UPDATE:感谢aynber,我已经跟踪误差。这是第一个查询错误:准备好的声明\“editRecord \”不存在”我不明白为什么这个工程的本地服务器上,但不能在预期的一个
更新2:之间错误准备的语句和执行:语法错误处或附近\ “ON \” \ n线段3:
case "editRecord":
$id = openPandoraBox(post("id"));
$tutorAbsence = post("tutorAbsence");
$clientAbsence = post("clientAbsence");
if($tutorAbsence == "1") {
if(post("tutor") != "0") {
// ------------this query does not work.-----------
$absUpsSql = "INSERT INTO tutorabsence(id, tutorid, reason)
VALUES ($1, $2, $3)
ON CONFLICT (id)
DO UPDATE SET tutorid=$2, reason=$3";
$absUpsPrep = pg_prepare($conn, 'editRecord', $absUpsSql);
$absUpsQry = pg_execute($conn, 'editRecord',
array($id, post("tutor"), post("tutorreason"))
);
} else {
$tutorAbsence = "0";
};
} else {
$absDelSql = "DELETE FROM tutorabsence WHERE id=$1";
$absDelPrep = pg_prepare($conn, 'absDel', $absDelSql);
$absDelQry = pg_execute($conn, 'absDel', array($id));
};
if($clientAbsence == "1"){
if(post("client") != "0") {
// ------------this query does not work.-----------
$absUpsSql = "INSERT INTO clientabsence(id, clientid, reason)
VALUES ($1, $2, $3)
ON CONFLICT (id)
DO UPDATE SET clientid=$2, reason=$3";
$absUpsPrep = pg_prepare($conn, 'absUps', $absUpsSql);
$absUpsQry = pg_execute($conn, 'absUps',
array($id, post("client"), post("clientreason"))
);
} else {
$clientAbsence = "0";
};
} else {
$absDelSql = "DELETE FROM clientabsence WHERE id=$1";
$absDelPrep = pg_prepare($conn, 'absDelOne', $absDelSql);
$absDelQry = pg_execute($conn, 'absDelOne', array($id));
};
$resultSql = "UPDATE appointments
SET hour=$1, tutorid=$2,
clientid=$3, purpose=$4,
tutornotshown=$5, clientnotshown=$6
WHERE appid=$7";
$resultPrep = pg_prepare($conn, 'resultSql', $resultSql);
$result = pg_execute($conn, 'resultSql',
array(post('hour'), post("tutor"), post("client"),
post("purpose"), $tutorAbsence, $clientAbsence, $id
)
);
echo json_encode(array("success" => 1));
break;
你检查了[错误](http://php.net/manual/en/function.pg-last-error.php)吗?什么是$ 1,$ 2和$ 3? – aynber
我无法从控制台收到任何错误。我在跟踪错误时遇到了问题,我是一个初学者。 $ 1,$ 2,$ 3是准备好的陈述的占位符。 – Ahmet
查看我发布的链接(点击单词错误),它会帮助你检查错误。 – aynber