Question

我有一个JavaScript函数，可以调用图片库。我使用php在我的数据库中获取img的名称，并尝试将结果作为参数传递给js函数，因此它将在我的页面上加载imgs。它应该加载4个图像，但是它假定值<img src="img/array1array2array3array4" />，它应该是<img src="img/array1" />，<img src="img/array2" />等等......我怎么能这样做？这是我的JS功能：

function galerias(query){


        var imgs = document.querySelector("#gallery");
        var x = 1;

            imgs.innerHTML += "<div class='row'>";
            imgs.innerHTML += "<div class='eight columns'>";
            imgs.innerHTML += "<h4>Galeria "+x+"</h4>"

        for(var i = 0; i < query.length; i++){
            imgs.innerHTML += "<img src='img/"+query[i]+"'class='imgs-galeria'>";

             x++;
        }

        imgs.innerHTML += "</div>";
        imgs.innerHTML += "</div>";
        imgs.innerHTML += "<a class='row' href='pics.html?gal="+x+"'><div class='twelve columns link'><p>Veja mais</p></div>";

}

PHP数据库查询：

function consultarDados($query){
    $dbResult = array();
    $conexao = mysql_connect($this->host, $this->usuario, $this->senha);
    mysql_select_db($this->banco, $conexao);
    $rs = mysql_query($query, $conexao);
        while($rows = mysql_fetch_assoc($rs)){
            //array push insere valores do $rows ao final do array $dbResult
            array_push($dbResult, $rows);
        }
    return $dbResult;
    mysql_close($conexao);
}

这是我在图片页面中进行查询的方式：

include 'connectDB.php';
$conexao = new connectDb();
$galeria_1= $conexao->consultarDados('select * from portfolio where gal= 1 and theme= 1');

$parametroGal = array();
foreach($galeria_1 as $result){

        array_push($parametroGal, $result['nome']);
    }

我如何将值作为参数传递给我的window.onload上的JS函数：

<?php
        echo'galerias("'.$parametroGal.'");';
?>

结果：

<div class="eight columns">
    //that way it just give me as result the word array, since $parametroGal is an array
    <img src="img/A" class="imgs-galeria">
    <img src="img/r" class="imgs-galeria">
    <img src="img/r" class="imgs-galeria">
    <img src="img/a" class="imgs-galeria">
    <img src="img/y" class="imgs-galeria">
</div>

//i need to get the value of that array, like this:

<div class="eight columns">
    //the result of the array, the results of the query in the DB
    <img src="img/foto1_1.jpg" class="imgs-galeria">
    <img src="img/foto1_2.jpg" class="imgs-galeria">
    <img src="img/foto1_3.jpg" class="imgs-galeria">
    <img src="img/foto1_4.jpg" class="imgs-galeria">
</div>

Answer 1

你有没有试过这样的事情：

<?php
        echo'<script>galerias("'.implode("\",\"",$parametroGal).'");</script>';
?>

<强>被修改

将数组作为参数传递但不起作用

1 个答案: