Mark Rotteveel - 感谢您的建议。
我正在使用 bindParam 方法来保存数据。
我通过使用以下 javascript 代码解决了它
function clk_submit(){
//document.getElementById("loader").style.display = "inline";
var str = document.getElementById("id_start_time").value;
var s_time = str.replace("T", " ");
var str = document.getElementById("id_end_time").value;
var e_time = str.replace("T", " ");
var xdata = {
bpr_pid:document.getElementById("id_product_id").value,
bpr_bno:document.getElementById("id_batch_no").value,
bpr_stageid:document.getElementById("id_stage").value,
bpr_cpid:"0",
bpr_bwt:document.getElementById("id_batch_weight").value,
bpr_status_id:"0",
bpr_stime:s_time,
bpr_etime:e_time,
bpr_closed:document.getElementById("id_closed").value,
bpr_uid:"<?php echo $_SESSION["userid"] ?>",
bpr_rem:"",
bpr_inbid:"0",
bpr_fcid:document.getElementById("id_formula_code").value,
bpr_seq_code_id:document.getElementById("id_sequence_code").value
};
var myJSON = JSON.stringify(xdata);
var xmlhttp;
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
// document.getElementById("loader").style.display = "none";
if (this.responseText==1) {
alert( "Record Saved Successfully");
} else {
alert(this.responseText);
}
}
};
xmlhttp.open("POST", "json_save_to_db.php", true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.send("x=" + myJSON);
}
现在完美了。