Question

I know this may be a duplicate, but I cant wrap my brain around the other examples. Help would be appreciated. I have a php array that i need to assign to a javascript array. Here is my amateur way of doing it now. You can see source at http://www.preferweb.com/accentps/index.php

<?php

$i=0;
while ($result1 = mysql_fetch_array($query1)){
print "<script>";
print "var size[".$i."]=" .$result1[ type ].";
";
print "var 25[".$i."]=" .$result1[ 25 ].";
";
print "var 50[".$i."]=" .$result1[ 50 ].";
";
print "var 100[".$i."]=" .$result1[ 100 ].";
";
print "var 250[".$i."]=" .$result1[ 250 ].";
";
print "var 500[".$i."]=" .$result1[ 500 ].";
";
print "var plus[".$i."]=" .$result1[ plus ].";
";
$i = $i+1;
}
print "var tick= 1 ;
";
print "alert (tick);
";
print "</script>
";
?>
<script>
alert (500[0]);

</script>

这种警示没有为警示和第二次警示作出任何规定。感谢。

Answer 1

You cannot use an integer as a variable name, like in this line: print "var 25[".$i."]=" .$result1[ 25 ]."; ";. 25 cannot be a variable.

如果你想把一个阵列描绘成一个javascript物体,你可能会想看json_encode 。

EXAMPLE
Your code could be written like this:

<?php
$result = array();

while ($row = mysql_fetch_array($query1)){
  $result[] = $row;
}
?>
<script>
  var result = <?= json_encode($result); ?>;
  alert (result[1][500]);
</script>

看起来对我来说更清洁。

Answer 2

你与阵列合作的方式是不正确的。

首先,你应首先确定阵列:

var myArr = [];

那么,如果你只是想增加阵列,你可以推动:

myArr.push("something");

或具体指数:

myArr[11] = "something";

你使用的辛子完全无效。

Answer 3

你的代码是错误的,因为购买力平价会产生什么(特别是因为你把数字作为瓦贾文的变数,加上你对每一处的相同变量)。

为了简化你想要达到的目标,仅就在购买力平价中产生一些变数,并为之分配价值。引自:$my_proxy_var。

之后, 绕过 Java的<>/strong>,如上(在<script> tag):

var myProxyVar = <?php echo json_encode($my_proxy_var); ?>;

有理由记住:

non-associative array in PHP becomes simple array in JavaScript,

associative array in PHP becomes object in JavaScript,

这一点很重要,因此,你可以避免混乱,选择各级非联系和联系阵列。
可在 

Answer 4

You can t use numbers as variable names in javascript.

你们不需要使用每一行的“老”。类似于

var test = [];
test[1] =  some value ;
test[2] =  some value ;

或许,请看一下使用 JSON_ENCODE从PHP的功能。

Answer 5

<?php
    if (!func_exists( json_encode )) die( sorry... I tried );
    $buffer = array();
    while ($value = mysql_fetch_assoc($result)) {
        $buffer[] = $value;
    }
    echo "<script>var data = ".json_encode($buffer)."</script>";
?>
<script>
console.log(data);
</script>

所需PHP 5.2.0

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