Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/230.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Javascript 使用PHP将SQL值获取到jsPDF中_Javascript_Php_Sql_Jspdf - Fatal编程技术网

Javascript 使用PHP将SQL值获取到jsPDF中

Javascript 使用PHP将SQL值获取到jsPDF中,javascript,php,sql,jspdf,Javascript,Php,Sql,Jspdf,我目前正在开发一个发票系统,并试图将发票导出为PDF格式。PDF创建应该使用jsPDF完成,发票的值(客户、地址、订单号等)在SQL数据库中。 我的计划是,创建PDF文件的.js程序从另一个PHP文档获取该文档 var AnRD = <?php include "GetData.php"; return json_encode(giveAnRD(ResID), JSON_HEX_TAG);?>;; var AbRD = <?php include "GetD

我目前正在开发一个发票系统,并试图将发票导出为PDF格式。PDF创建应该使用jsPDF完成,发票的值(客户、地址、订单号等)在SQL数据库中。 我的计划是,创建PDF文件的.js程序从另一个PHP文档获取该文档

var AnRD =      <?php include "GetData.php"; return json_encode(giveAnRD(ResID), JSON_HEX_TAG);?>;;
var AbRD =      <?php include "GetData.php"; return json_encode(giveAbRD(ResID), JSON_HEX_TAG);?>;;
var VName =     <?php include "GetData.php"; return json_encode(giveVorname(ResID), JSON_HEX_TAG);?>;;
var NName =     <?php include "GetData.php"; return json_encode(giveNachname(ResID), JSON_HEX_TAG);?>;;
var Strasse =   <?php include "GetData.php"; return json_encode(giveStarsse(ResID), JSON_HEX_TAG);?>;;
var Hnummer =   <?php include "GetData.php"; return json_encode(giveHnummer(ResID), JSON_HEX_TAG);?>;;
var PLZ =       <?php include "GetData.php"; return json_encode(givePLZ(ResID), JSON_HEX_TAG);?>;;
var Ort =       <?php include "GetData.php"; return json_encode(giveOrt(ResID), JSON_HEX_TAG);?>;;
var Anrede =    <?php include "GetData.php"; return json_encode(giveAn(ResID), JSON_HEX_TAG);?>;;
你知道为什么这样不行吗

提前谢谢


docunhown

1)它怎么不起作用?;2) 向我们展示您试图追踪问题的步骤。3)在PHP中,如果您双引号引一个字符串并在其中插入有效变量(例如
$PID
),它将创建一个变量,并且该变量将被引用的变量的值替换,因此如果我们有
$foo='bar'$foobar=“foo$bar”
,然后
$foobar
变成
foobar
;4) SQL错误:
WHERE PersonID AND NOT NULL='SomeValue'
毫无意义。您的意思是
其中PersonID='SomeValue'
?@tyteen4a03与您提到的SQL错误有关:
和不为NULL
放在那里,因为字段
Anrede
可能为NULL,但我从代码中删除了它。谢谢你的建议。To 1):通常,当我执行代码js代码时,应该执行一个PDF文件。但是,在我将字段替换为例如名称的var之后,我没有从文件中得到任何响应;文件文本(20,82,PLZ+“”+Ort)
include "../dblogin.php";
$resID="";


$db = new PDO("mysql:host=$dbhost;dbname=$dbname;", "$dbuser", "$dbpass");
//##############################################
function giveAnRD($resID)
{
    $AnRD = "SELECT reserviert_von FROM reservierungen WHERE ReservierungsID = $resID";
    return $AnRD;
}
//##############################################
function giveHausName($resID)
{
    $HID = "SELECT HausID FROM reservierungen WHERE ReservierungsID = $resID";
    $AnRD = "SELECT Hausname FROM hauser WHERE HausID = $HID";
    return $AnRD;
}
//##############################################
function giveAbRD($resID)
{
    $AnRD = "SELECT reserviert_bis FROM reservierungen WHERE ReservierungsID = $resID";
    return $AnRD;
}
//##############################################
function giveAnrede($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";

    if ($TITEL =! NULL)
    {
        $Anrede = $GEN . ' ' . $TITEL;
    }

    else
    {
        $Anrede = $GEN;
    }
$GEN = "SELECT Anrede FROM persons WHERE PersonID = $PID";
return $GEN;
}
//##############################################
function giveVorname($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}
//##############################################
function giveNachname($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}
//##############################################
function giveStrasse($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}
//##############################################
function giveHnummer($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}
//##############################################
function givePLZ($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}
//##############################################
function giveOrt($resID)
{
    $PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
    $TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
    return $TITEL;
}