如何将PHP数组转换成这样的格式
Array
(
[0] => 001-1234567
[1] => 1234567
[2] => 12345678
[3] => 12345678
[4] => 12345678
[5] => AP1W3242
[6] => AP7X1234
[7] => AS1234
[8] => MH9Z2324
[9] => MX1234
[10] => TN1A3242
[11] => ZZ1234
)
到下面格式的Javascript数组?
var cities = [
"Aberdeen",
"Ada",
"Adamsville",
"Addyston",
"Adelphi",
"Adena",
"Adrian",
"Akron",
"Albany"
];
当前回答
对于Laravel用户:使用@json
var cities = @json($data);
官方文档中有解释:https://laravel.com/docs/8.x/blade#rendering-json:
其他回答
很简单，我用这个方法:
JAVASCRIPT:
var arrayJavascript = <?php echo json_encode($arrayPhp) ?>;
转换PHP数组到Javascript
的问候!
愚蠢而简单:
var js_array = [<?php echo '"'.implode('","', $php_array).'"' ?>];
这是我的函数。JavaScript必须在PHP下，否则使用SESSION。
<?php
$phpArray=array(1,2,3,4,5,6);
?>
<div id="arrayCon" style="width:300px;"></div>
<script type="text/javascript">
var jsArray = new Array();
<?php
$numArray=count($phpArray);
for($i=0;$i<$numArray;$i++){
echo "jsArray[$i] = ". $phpArray[$i] . ";\n";
}
?>
$("#arrayCon").text(jsArray[1]);
</script>
最后一行可以是....text(jsArray);会显示"1,2,3,4,5,6"
我使用了一个伪php数组
<?php
// instead to create your array like this
$php_array = ["The","quick","brown","fox","jumps","over","the","lazy","dog"];
// do it like this (a simple variable but with separator)
$php_fake_array = "The,quick,brown,fox,jumps,over,the,lazy,dog";
?>
<script type="text/javascript">
// use the same separator for the JS split() function
js_array = '<?php echo $php_fake_array; ?>'.split(',');
</script>
如果你的数组是未知的(已经创建)
<?php
$php_array = file('my_file.txt');
$php_fake_array = "";
// transform your array with concatenate like this
foreach ($php_array as $cell){
// since this array is unknown, use clever separator
$php_fake_array .= $cell.",,,,,";
}
?>
<script type="text/javascript">
// use the same separator for the JS split() function
js_array = '<?php echo $php_fake_array; ?>'.split(',,,,,');
</script>
可以用更安全的方式来做。
如果你的PHP数组包含特殊字符，你需要在PHP中使用rawurlencode()，然后在JS中使用decodeURIComponent()来转义这些字符。并将JSON解析为原生js。试试这个:
var data = JSON.parse(
decodeURIComponent(
"<?=rawurlencode(json_encode($data));?>"
)
);
console.log(data);
