如何将变量从PHP传递到生成的JavaScript？

我正在从PHP脚本生成JavaScript，我需要能够将值直接复制到生成的JavaScript代码中

我的代码如下：

PHP:

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

---

问题是：

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

行

$price=$\u POST[price]

产生一个未定义的值

我试图将

$price

复制到PHP脚本中，但它在我的脚本中逐字输出

$price

而不是

$price

中包含的值

---

问题1:

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

您需要将

$\u POST[price]

更改为

$\u POST['price']

，就像您刚才在上面所做的那样。如果不使用引号将其括起来，PHP将假定您正在尝试使用通过

define（）

定义的常量

---

问题2:

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

如果您使用的是香草PHP，并且希望向脚本中注入变量，您可以这样做：

var price = ($this).val(<?php echo json_encode($price); ?>);

var price=（$this）.val（）；

当然，这假设JavaScript是从上面的PHP脚本生成的

您可以阅读文档以了解更多信息

---

您的代码中也存在巨大的安全漏洞：

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

目前，您的脚本直接从用户请求中获取数据，并将其放入SQL查询中。这可能导致SQL注入攻击。不太好

可以找到关于这个主题的官方PHP文档

---

建议：

include "db-Info.php";
echo '<option value="">Please select item first...</option>';

$item = $_POST['itemId'];

$query = mysqli_query($con, "SELECT id, item, price, pointRequired FROM 
tblprice WHERE item ='$item'");
$rowCount = $query->num_rows;

if($rowCount > 0) {
    while($row = mysqli_fetch_array($query)){
        $price = $row['price']; 
        $pointRequired = $row['pointRequired'];

        echo "<option id='priceNew'>";
        echo "Price : $price" ?>&nbsp; <?php echo "Point: $pointRequired";
        echo "</option>";      
    }

    if($item){  
        $price = $_POST[price];//shows undefined 
        $query1 = mysqli_query($con, "SELECT id FROM tblprice WHERE item 
='$item'AND price ='$price'"); //it seems not picking up the same id with 
the item I selected above
        $rows = mysqli_fetch_array($query1);
        $id = (string) reset($rows);
        $barcodeNew = $id . $item;

        echo $barcodeNew;
    }
}

$(document).ready(function() {
$('#price-select').on('change', function(){
    var price = $(this).val($price); // This is where $price needs to copy
    if(price){
        $.ajax({
            type:'POST',
            url: 'functions/pos-getPrice.php',
            data: {price:price},  
            success:function(html){ 
                alert(price);
            }
        })
    }
 })
})//THIS CODE BLOCK IS PROBLEM 

首先，我注意到你在全球范围内做一切事情。我强烈建议您学习如何将代码封装在类和函数中；这将使维护和测试变得更加容易

可以让您走上正确道路的好东西有：

• PHP面向对象编程（OOP）
• PHP标准建议（PSR）
• PHP编写器

第二，这样做很过时，很容易遇到问题。我鼓励您研究更现代的框架，例如。您还可以查找单页应用程序（SPA）框架，如、或，它们与作为后端RESTAPI的Laravel配合使用非常好

---

祝你好运。

你能不能在打开jqery的情况下，在页面上回显该值

var foo=

我尝试并测试了警报，但它显示了代码本身“我不知道为什么”